इयत्ता 8 वी विस्तार सूत्रे म्हणजे काय? : विस्तार सूत्रे म्हणजे बहुपदीय समीकरणे किंवा गृहितांकांच्या स्वरूपात दिलेल्या संकल्पनांचे विस्तारित स्वरूप. उदाहरणार्थ, ( a + b ) 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+2ab+b^2 ( a + b ) 2 = a 2 + 2 a b + b 2
विस्तार सूत्रे म्हणजे काय?
विस्तार सूत्रे ही गणितातील महत्त्वाची संकल्पना असून ती बहुपदींच्या विस्तारासाठी आणि सरलीकरणासाठी वापरली जातात. विस्तार सूत्रे बीजगणित आणि संख्याशास्त्रात महत्त्वाची भूमिका बजावतात. संगणकीय गणितीय मॉडेलिंगसाठी आणि अल्गोरिदमच्या अचूकतेसाठी देखील विस्तार सूत्रे उपयोगी पडतात.
मूलभूत विस्तार सूत्रे
( x + y ) 2 = x 2 + 2 x y + y 2 (x+y)^2 = x^{2} + 2xy + y^{2} ( x + y ) 2 = x 2 + 2 x y + y 2 ( x − y ) 2 = x 2 − 2 x y + y 2 (x-y)^2 = x^2 - 2xy + y^2 ( x − y ) 2 = x 2 − 2 x y + y 2 ( x + y ) ( x − y ) = x 2 − x y + x y − y 2 = x 2 − y 2 (x+y)(x-y) = x^2 - xy + xy - y^2 = x^2 - y^2 ( x + y ) ( x − y ) = x 2 − x y + x y − y 2 = x 2 − y 2
मित्रांनो, ह्या वरून विस्तार सूत्रे म्हणजे काय, हे तुमच्या लक्षात आलं असेल. ही विस्तार सूत्रे आणि आपण पुढे जी शिकणार आहोत ती सगळी विस्तार सूत्रे पक्की लक्षात ठेवण्यासाठी तुम्ही ह्या विस्तार सूत्रांचा एक तक्ता तयार करून तुमच्या घरात भिंतीवर लावून ठेवा आणि जाता येता त्यावर नजर टाका, जेणे करून ही विस्तार सूत्रे तुमच्या कायम लक्षात राहतील.
चला तर मग आता आपण ह्या विस्तारसुत्रांची काही उदाहरणं सोडवून बघूया,
उदाहरण 1: ( x + 2 y ) 2 (x+2y)^2 ( x + 2 y ) 2 उत्तर: ( x + 2 y ) 2 = x 2 + 4 x y + 4 y 2 (x+2y)^2 = x^2 + 4xy + 4y^2 ( x + 2 y ) 2 = x 2 + 4 x y + 4 y 2
उदाहरण 2: ( 2 x − 5 y ) 2 (2x-5y)^2 ( 2 x − 5 y ) 2 उत्तर: ( 2 x − 5 y ) 2 = ( 2 x ) 2 − 2 ( 10 x y ) + ( 5 y ) 2 = 4 x 2 − 20 x y + 25 y 2 \begin{aligned}(2x-5y)^2 &= (2x)^2 - 2(10xy) + (5y)^2 \\ &= 4x^2 - 20xy + 25y^2\end{aligned} ( 2 x − 5 y ) 2 = ( 2 x ) 2 − 2 ( 1 0 x y ) + ( 5 y ) 2 = 4 x 2 − 2 0 x y + 2 5 y 2
उदाहरण 3: ( 101 ) 2 (101)^2 ( 1 0 1 ) 2 उत्तर: ( 101 ) 2 = ( 100 + 1 ) 2 = ( 100 ) 2 + 2 ( 100 × 1 ) + 1 2 = 10000 + 200 + 1 = 10201 \begin{aligned} \\ (101)^2 &= (100+1)^2\\ &=(100)^2+2(100\times 1)+1^2 \\ &=10000+200+1 \\ &=10201\end{aligned} ( 1 0 1 ) 2 = ( 1 0 0 + 1 ) 2 = ( 1 0 0 ) 2 + 2 ( 1 0 0 × 1 ) + 1 2 = 1 0 0 0 0 + 2 0 0 + 1 = 1 0 2 0 1
उदाहरण 4: ( 98 ) 2 (98)^2 ( 9 8 ) 2 उत्तर: ( 98 ) 2 = ( 100 − 2 ) 2 = ( 100 ) 2 − 2 ( 100 × 2 ) + ( 2 ) 2 = 10000 − 400 + 4 = 9604 \begin{aligned} \\ (98)^2 &= (100-2)^2 \\ &=(100)^2-2(100\times 2)+(2)^2\\&=10000-400+4\\&=9604\end{aligned} ( 9 8 ) 2 = ( 1 0 0 − 2 ) 2 = ( 1 0 0 ) 2 − 2 ( 1 0 0 × 2 ) + ( 2 ) 2 = 1 0 0 0 0 − 4 0 0 + 4 = 9 6 0 4
उदाहरण 5: ( 5 m + 3 n ) ( 5 m − 3 n ) (5m+3n)(5m-3n) ( 5 m + 3 n ) ( 5 m − 3 n ) उत्तर: ( 5 m + 3 n ) ( 5 m − 3 n ) = ( 5 m ) 2 − ( 3 n ) 2 = 25 m 2 − 9 n 2 \begin{aligned} \\ (5m+3n)(5m-3n) &= (5m)^2-(3n)^2 \\ &=25m^2-9n^2\end{aligned} ( 5 m + 3 n ) ( 5 m − 3 n ) = ( 5 m ) 2 − ( 3 n ) 2 = 2 5 m 2 − 9 n 2
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इयत्ता 8 वी विस्तार सूत्रे म्हणजे काय? हे समजावून घेताना आयत आणि चौरस यांच्या क्षेत्रफळांच्या सूत्रांच्या साहाय्याने ( x + a ) ( x + b ) \mathbf{(x+a)(x+b)} ( x + a ) ( x + b ) चा विस्तार …
चौरसाचे आणि आयताचे क्षेत्रफळ काढायचे सूत्र हे एकच आहे.चौरसाचे / आयताचे क्षेत्रफळ = लांबी x रुंदी
पण इथे एक लक्षात घ्या की चौरसाच्या चारही बाजू ह्या समान असतात आणि त्यामुळे लांबी आणि रुंदी समान असते.∴ \therefore ∴ चौरसाचे क्षेत्रफळ = (चौरसाची कोणतीही बाजू)2
∴ ( x + a ) ( x + b ) = x 2 + a x + b x + a b \therefore (x+a)(x+b)=x^2+ ax+bx+ab ∴ ( x + a ) ( x + b ) = x 2 + a x + b x + a b ∴ ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b \therefore (x+a)(x+b)=x^2+ x(a+b)+ ab ∴ ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b
महत्वाचं: ( x + a ) \mathbf{(x+a)} ( x + a ) ( x + b ) \mathbf{(x+b)} ( x + b ) एक पद समान असलेल्या द्विपदी म्हणतात.( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b \mathbf{(x+a)(x+b)=x^2+ x(a+b)+ab} ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b
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विस्तार सूत्रांची उदाहरणे उदाहरण 1: ( x + 3 ) ( x + 5 ) (x+3)(x+5) ( x + 3 ) ( x + 5 ) उत्तर: ( x + 3 ) (x+3) ( x + 3 ) ( x + 5 ) (x+5) ( x + 5 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( x + 3 ) ( x + 5 ) = x 2 + 5 x + 3 x + ( 3 × 5 ) = x 2 + x ( 5 + 3 ) + ( 15 ) = x 2 + 8 x + 15 \begin{aligned} \\ \therefore (x+3)(x+5)&=x^2+5x+3x+(3\times 5) \\ &= x^2+x(5+3)+(15) \\ &= x^2+8x+15\end{aligned} ∴ ( x + 3 ) ( x + 5 ) = x 2 + 5 x + 3 x + ( 3 × 5 ) = x 2 + x ( 5 + 3 ) + ( 1 5 ) = x 2 + 8 x + 1 5
उदाहरण 2: ( x + 7 ) ( x − 2 ) (x+7)(x-2) ( x + 7 ) ( x − 2 ) उत्तर: ( x + 7 ) (x+7) ( x + 7 ) ( x − 2 ) (x-2) ( x − 2 ) ( x + a ) ( x − b ) = x 2 + x ( a + b ) + a b (x+a)(x-b)=x^2+ x(a+b)+ab ( x + a ) ( x − b ) = x 2 + x ( a + b ) + a b ∴ ( x + 7 ) ( x − 2 ) = x 2 − 2 x + 7 x + ( 7 × ( − 2 ) ) = x 2 + x ( − 2 + 7 ) + ( − 14 ) = x 2 + 5 x − 14 \begin{aligned} \\ \therefore (x+7)(x-2)&= x^2-2x+7x+(7\times (-2)) \\ &= x^2+x(-2+7)+(-14) \\ &=x^2+5x-14\end{aligned} ∴ ( x + 7 ) ( x − 2 ) = x 2 − 2 x + 7 x + ( 7 × ( − 2 ) ) = x 2 + x ( − 2 + 7 ) + ( − 1 4 ) = x 2 + 5 x − 1 4
उदाहरण 3: ( 2 a + 3 b ) ( 2 a − 3 b ) (2a+3b)(2a-3b) ( 2 a + 3 b ) ( 2 a − 3 b ) उत्तर: ( 2 a + 3 b ) (2a+3b) ( 2 a + 3 b ) ( 2 a − 3 b ) (2a-3b) ( 2 a − 3 b ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( 2 a + 3 b ) ( 2 a − 3 b ) = ( 2 a ) 2 − ( 2 a × 3 b ) + ( 3 b × 2 a ) − ( 3 b × 3 b ) = 4 a 2 + ( − 3 b + 3 b ) 2 a − 9 b 2 = 4 a 2 + ( 0 ) 2 a − 9 b 2 = 4 a 2 − 9 b 2 \begin{aligned} \\ \therefore (2a+3b)(2a-3b)&=(2a)^2-(2a\times3b)+(3b\times2a)-(3b\times3b) \\ &= 4a^2+(-3b+3b)2a-9b^2 \\ &= 4a^2+(0)2a-9b^2 \\ &= 4a^2-9b^2\end{aligned} ∴ ( 2 a + 3 b ) ( 2 a − 3 b ) = ( 2 a ) 2 − ( 2 a × 3 b ) + ( 3 b × 2 a ) − ( 3 b × 3 b ) = 4 a 2 + ( − 3 b + 3 b ) 2 a − 9 b 2 = 4 a 2 + ( 0 ) 2 a − 9 b 2 = 4 a 2 − 9 b 2
उदाहरण 4: ( m + 3 2 ) ( m + 1 2 ) \left(m+\frac{3}{2}\right)\left(m+\frac{1}{2}\right) ( m + 2 3 ) ( m + 2 1 ) उत्तर: ( m + 3 2 ) \left(m+\frac{3}{2}\right) ( m + 2 3 ) ( m + 1 2 ) \left(m+\frac{1}{2}\right) ( m + 2 1 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( m + 3 2 ) ( m + 1 2 ) = m 2 + ( 3 2 + 1 2 ) m + ( 3 2 × 1 2 ) = m 2 + ( 4 2 ) m + ( 3 4 ) = m 2 + 2 m + 3 4 \begin{aligned} \\ \therefore \left(m+\frac{3}{2}\right)\left(m+\frac{1}{2}\right)&=m^2+\left(\frac{3}{2}+\frac{1}{2}\right)m+\left(\frac{3}{2}\times \frac{1}{2}\right) \\ &=m^2+\left(\frac{4}{2}\right)m+\left(\frac{3}{4}\right) \\ &=m^2+2m+\frac{3}{4}\end{aligned} ∴ ( m + 2 3 ) ( m + 2 1 ) = m 2 + ( 2 3 + 2 1 ) m + ( 2 3 × 2 1 ) = m 2 + ( 2 4 ) m + ( 4 3 ) = m 2 + 2 m + 4 3
उदाहरण 5: ( x − 7 ) ( x − 2 ) \left(x-7\right)\left(x-2\right) ( x − 7 ) ( x − 2 ) उत्तर: ( x − 7 ) \left(x-7\right) ( x − 7 ) ( x − 2 ) \left(x-2\right) ( x − 2 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( x − 7 ) ( x − 2 ) = x 2 + ( − 2 x − 7 x ) + ( − 7 × − 2 ) = x 2 + ( − 2 − 7 ) x + 14 = x 2 − 9 x + 14 \begin{aligned} \\ \therefore \left(x-7\right)\left(x-2\right)&= x^2+\left(-2x-7x\right)+\left(-7\times -2\right) \\ &= x^2+\left(-2-7\right)x+14 \\ &= x^2-9x+14\end{aligned} ∴ ( x − 7 ) ( x − 2 ) = x 2 + ( − 2 x − 7 x ) + ( − 7 × − 2 ) = x 2 + ( − 2 − 7 ) x + 1 4 = x 2 − 9 x + 1 4
उदाहरण 6: ( a + 2 ) ( a − 1 ) \left(a+2\right)\left(a-1\right) ( a + 2 ) ( a − 1 ) उत्तर: ( a + 2 ) \left(a+2\right) ( a + 2 ) ( a − 1 ) \left(a-1\right) ( a − 1 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( a + 2 ) ( a − 1 ) = a 2 + ( − a + 2 a ) + ( 2 × − 1 ) = a 2 + a − 2 \begin{aligned} \\ \therefore \left(a+2\right)\left(a-1\right)&=a^2+\left(-a+2a\right)+\left(2\times -1\right) \\ &=a^2+a-2\end{aligned} ∴ ( a + 2 ) ( a − 1 ) = a 2 + ( − a + 2 a ) + ( 2 × − 1 ) = a 2 + a − 2
उदाहरण 7: ( m − 4 ) ( m + 6 ) \left(m-4\right)\left(m+6\right) ( m − 4 ) ( m + 6 ) उत्तर: ( m − 4 ) \left(m-4\right) ( m − 4 ) ( m + 6 ) \left(m+6\right) ( m + 6 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( m − 4 ) ( m + 6 ) = m 2 + ( 6 m − 4 m ) + ( − 4 × 6 ) = m 2 + 2 m − 24 \begin{aligned} \\ \therefore \left(m-4\right)\left(m+6\right)&=m^2+\left(6m-4m\right)+\left(-4\times 6\right) \\ &=m^2+2m-24\end{aligned} ∴ ( m − 4 ) ( m + 6 ) = m 2 + ( 6 m − 4 m ) + ( − 4 × 6 ) = m 2 + 2 m − 2 4
उदाहरण 8: ( p + 8 ) ( p − 3 ) \left(p+8\right)\left(p-3\right) ( p + 8 ) ( p − 3 ) उत्तर: ( p + 8 ) \left(p+8\right) ( p + 8 ) ( p − 3 ) \left(p-3\right) ( p − 3 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( p + 8 ) ( p − 3 ) = p 2 + ( − 3 p + 8 p ) + ( 8 × − 3 ) = p 2 + 5 p − 24 \begin{aligned} \\ \therefore\left(p+8\right)\left(p-3\right)&=p^2+\left(-3p+8p\right)+\left(8\times -3\right) \\ &=p^2+5p-24\end{aligned} ∴ ( p + 8 ) ( p − 3 ) = p 2 + ( − 3 p + 8 p ) + ( 8 × − 3 ) = p 2 + 5 p − 2 4
उदाहरण 9: ( a + 2 ) ( a − 1 ) \left(a+2\right)\left(a-1\right) ( a + 2 ) ( a − 1 ) उत्तर: ( a + 2 ) \left(a+2\right) ( a + 2 ) ( a − 1 ) \left(a-1\right) ( a − 1 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( a + 2 ) ( a − 1 ) = a 2 − a + 2 a − 2 = a 2 + a − 2 \begin{aligned} \\ \therefore \left(a+2\right)\left(a-1\right)&=a^2-a+2a-2 \\ &=a^2+a-2\end{aligned} ∴ ( a + 2 ) ( a − 1 ) = a 2 − a + 2 a − 2 = a 2 + a − 2
उदाहरण 10: ( p + 8 ) ( p − 3 ) \left(p+8\right)\left(p-3\right) ( p + 8 ) ( p − 3 ) उत्तर: ( p + 8 ) \left(p+8\right) ( p + 8 ) ( p − 3 ) \left(p-3\right) ( p − 3 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( p + 8 ) ( p − 3 ) = p 2 − 3 p + 8 p − 24 = p 2 + 5 p − 24 \begin{aligned} \\ \therefore \left(p+8\right)\left(p-3\right)&=p^2-3p+8p-24 \\ &=p^2+5p-24\end{aligned} ∴ ( p + 8 ) ( p − 3 ) = p 2 − 3 p + 8 p − 2 4 = p 2 + 5 p − 2 4
उदाहरण 11: ( m − 4 ) ( m + 6 ) \left(m-4\right)\left(m+6\right) ( m − 4 ) ( m + 6 ) उत्तर: ( m − 4 ) \left(m-4\right) ( m − 4 ) ( m + 6 ) \left(m+6\right) ( m + 6 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( m − 4 ) ( m + 6 ) = m 2 + 6 m − 4 m − 24 = m 2 + 2 m − 24 \begin{aligned} \\ \therefore \left(m-4\right)\left(m+6\right)&=m^2+6m-4m-24 \\ &=m^2+2m-24\end{aligned} ∴ ( m − 4 ) ( m + 6 ) = m 2 + 6 m − 4 m − 2 4 = m 2 + 2 m − 2 4
उदाहरण 12: ( 13 + x ) ( 13 − x ) \left(13+x\right)\left(13-x\right) ( 1 3 + x ) ( 1 3 − x ) उत्तर: ( 13 + x ) \left(13+x\right) ( 1 3 + x ) ( 13 − x ) \left(13-x\right) ( 1 3 − x ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( 13 + x ) ( 13 − x ) = 1 3 2 − 13 x + 13 x − x 2 = 169 − x 2 \begin{aligned} \\ \therefore \left(13+x\right)\left(13-x\right)&=13^2-13x+13x-x^2 \\ &=169-x^2\end{aligned} ∴ ( 1 3 + x ) ( 1 3 − x ) = 1 3 2 − 1 3 x + 1 3 x − x 2 = 1 6 9 − x 2
उदाहरण 13: ( 3 x + 4 y ) ( 3 x + 5 y ) \left(3x+4y\right)\left(3x+5y\right) ( 3 x + 4 y ) ( 3 x + 5 y ) उत्तर: ( 3 x + 4 y ) \left(3x+4y\right) ( 3 x + 4 y ) ( 3 x + 5 y ) \left(3x+5y\right) ( 3 x + 5 y ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( 3 x + 4 y ) ( 3 x + 5 y ) = ( 3 x × 3 x ) + ( 3 x × 5 y ) + ( 4 y × 3 x ) + ( 4 y × 5 y ) = 9 x 2 + 15 x y + 12 x y + 20 y 2 = 9 x 2 + 27 x y + 20 y 2 \begin{aligned} \\ \therefore \left(3x+4y\right)\left(3x+5y\right)&=\left(3x\times 3x\right)+\left(3x\times 5y\right)+\left(4y\times 3x\right)+\left(4y\times 5y\right) \\ &=9x^2+15xy+12xy+20y^2 \\ &=9x^2+27xy+20y^2\end{aligned} ∴ ( 3 x + 4 y ) ( 3 x + 5 y ) = ( 3 x × 3 x ) + ( 3 x × 5 y ) + ( 4 y × 3 x ) + ( 4 y × 5 y ) = 9 x 2 + 1 5 x y + 1 2 x y + 2 0 y 2 = 9 x 2 + 2 7 x y + 2 0 y 2
उदाहरण 14: ( 9 x − 5 t ) ( 9 x + 3 t ) \left(9x-5t\right)\left(9x+3t\right) ( 9 x − 5 t ) ( 9 x + 3 t ) उत्तर: ( 9 x − 5 t ) \left(9x-5t\right) ( 9 x − 5 t ) ( 9 x + 3 t ) \left(9x+3t\right) ( 9 x + 3 t ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( 3 x + 4 y ) ( 3 x + 5 y ) = ( 3 x × 3 x ) + ( 3 x × 5 y ) + ( 4 y × 3 x ) + ( 4 y × 5 y ) = 9 x 2 + 15 x y + 12 x y + 20 y 2 = 9 x 2 + 27 x y + 20 y 2 \begin{aligned} \\ \therefore \left(3x+4y\right)\left(3x+5y\right)&=\left(3x\times3x\right)+\left(3x\times5y\right)+\left(4y\times3x\right)+\left(4y\times5y\right) \\ &=9x^2+15xy+12xy+20y^2 \\ &=9x^2+27xy+20y^2\end{aligned} ∴ ( 3 x + 4 y ) ( 3 x + 5 y ) = ( 3 x × 3 x ) + ( 3 x × 5 y ) + ( 4 y × 3 x ) + ( 4 y × 5 y ) = 9 x 2 + 1 5 x y + 1 2 x y + 2 0 y 2 = 9 x 2 + 2 7 x y + 2 0 y 2
उदाहरण 15: ( m + 2 3 ) ( m − 7 3 ) \left(m+\frac{2}{3}\right)\left(m-\frac{7}{3}\right) ( m + 3 2 ) ( m − 3 7 ) उत्तर: ( m + 2 3 ) \left(m+\frac{2}{3}\right) ( m + 3 2 ) ( m − 7 3 ) \left(m-\frac{7}{3}\right) ( m − 3 7 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( m + 2 3 ) ( m − 7 3 ) = ( m × m ) + ( − 7 3 m ) + ( 2 3 m ) + ( 2 3 × − 7 3 ) = m 2 + ( − 7 3 + 2 3 ) m − 14 9 = m 2 − 5 3 m − 14 9 \begin{aligned} \\ \therefore \left(m+\frac{2}{3}\right)\left(m-\frac{7}{3}\right)&=\left(m\times m\right)+\left(-\frac{7}{3} m\right)+\left(\frac{2}{3} m\right)+\left(\frac{2}{3}\times -\frac{7}{3}\right) \\ &=m^2+\left(-\frac{7}{3}+\frac{2}{3}\right)m-\frac{14}{9} \\ &=m^2-\frac{5}{3} m-\frac{14}{9}\end{aligned} ∴ ( m + 3 2 ) ( m − 3 7 ) = ( m × m ) + ( − 3 7 m ) + ( 3 2 m ) + ( 3 2 × − 3 7 ) = m 2 + ( − 3 7 + 3 2 ) m − 9 1 4 = m 2 − 3 5 m − 9 1 4
उदाहरण 16: ( x + 1 x ) ( x − 1 x ) \left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right) ( x + x 1 ) ( x − x 1 ) उत्तर: ( x + 1 x ) \left(x+\frac{1}{x}\right) ( x + x 1 ) ( x − 1 x ) \left(x-\frac{1}{x}\right) ( x − x 1 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( x + 1 x ) ( x − 1 x ) = ( x × x ) + ( − x x ) + ( x x ) + ( 1 x × − 1 x ) = x 2 − 1 + 1 − 1 x 2 = x 2 − 1 x 2 \begin{aligned} \\ \therefore \left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)&=\left(x\times x\right)+\left(-\frac{x}{x}\right)+\left(\frac{x}{x}\right)+\left(\frac{1}{x}\times -\frac{1}{x}\right) \\ &=x^2-1+1- \frac{1}{x^2} \\ &=x^2-\frac{1}{x^2}\end{aligned} ∴ ( x + x 1 ) ( x − x 1 ) = ( x × x ) + ( − x x ) + ( x x ) + ( x 1 × − x 1 ) = x 2 − 1 + 1 − x 2 1 = x 2 − x 2 1
उदाहरण 17: ( 1 y + 4 ) ( 1 y − 9 ) \left(\frac{1}{y}+4\right)\left(\frac{1}{y}-9\right) ( y 1 + 4 ) ( y 1 − 9 ) उत्तर: ( 1 y + 4 ) \left(\frac{1}{y}+4\right) ( y 1 + 4 ) ( 1 y − 9 ) \left(\frac{1}{y}-9\right) ( y 1 − 9 ) ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b (x+a)(x+b)=x^2+ x(a+b)+ab ( x + a ) ( x + b ) = x 2 + x ( a + b ) + a b ∴ ( 1 y + 4 ) ( 1 y − 9 ) = ( 1 y × 1 y ) + ( 1 y × − 9 ) + ( 4 × 1 y ) + ( 4 × − 9 ) = ( 1 y 2 ) − 9 y + 4 y − 36 = 1 y 2 − 5 y − 36 \begin{aligned} \\ \therefore \left(\frac{1}{y}+4\right)\left(\frac{1}{y}-9\right)&=\left(\frac{1}{y}\times \frac{1}{y}\right)+\left(\frac{1}{y}\times -9\right)+\left(4\times \frac{1}{y}\right)+\left(4\times -9\right) \\ &=\left(\frac{1}{y^2} \right)-\frac{9}{y}+\frac{4}{y}-36 \\ &=\frac{1}{y^2}-\frac{5}{y}-36\end{aligned} ∴ ( y 1 + 4 ) ( y 1 − 9 ) = ( y 1 × y 1 ) + ( y 1 × − 9 ) + ( 4 × y 1 ) + ( 4 × − 9 ) = ( y 2 1 ) − y 9 + y 4 − 3 6 = y 2 1 − y 5 − 3 6
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( a + b ) 3 \mathbf{\left(a+b\right)^3} ( a + b ) 3 चा विस्तार ( a + b ) 3 = ( a + b ) ( a + b ) 2 = ( a + b ) ( a 2 + 2 a b + b 2 ) = a ( a 2 + 2 a b + b 2 ) + b ( a 2 + 2 a b + b 2 ) = a 3 + 2 a 2 b + a b 2 + a 2 b + 2 a b 2 + b 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 \begin{aligned}\left(a+b\right)^3&=\left(a+b\right)\left(a+b\right)^2\\&=\left(a+b\right)\left(a^2+2ab+b^2\right)\\&=a\left(a^2+2ab+b^2 \right)+b\left(a^2+2ab+b^2\right)\\&=a^3+2a^2 b+ab^2+a^2 b+2ab^2+b^3\\&=a^3+3a^2 b+3ab^2+b^3\end{aligned} ( a + b ) 3 = ( a + b ) ( a + b ) 2 = ( a + b ) ( a 2 + 2 a b + b 2 ) = a ( a 2 + 2 a b + b 2 ) + b ( a 2 + 2 a b + b 2 ) = a 3 + 2 a 2 b + a b 2 + a 2 b + 2 a b 2 + b 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3
∴ ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 \mathbf{\therefore \left(a+b\right)^3=a^3+3a^2 b+3ab^2+b^3} ∴ ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3
( a + b ) 3 \left(a+b\right)^3 ( a + b ) 3
उदाहरण 1: ( x + 3 ) 3 = ( x + 3 ) ( x + 3 ) 2 = ( x + 3 ) ( x 2 + 6 x + 9 ) = x ( x 2 + 6 x + 9 ) + 3 ( x 2 + 6 x + 9 ) = x 3 + 6 x 2 + 9 x + 3 x 2 + 18 x + 27 = x 3 + 9 x 2 + 27 x + 27 \begin{aligned} \\ \left(x+3\right)^3&=\left(x+3\right)\left(x+3\right)^2 \\ &=\left(x+3\right)\left(x^2+6x+9\right) \\ &=x\left(x^2+6x+9\right)+3\left(x^2+6x+9\right) \\ &=x^3+6x^2+9x+3x^2+18x+27 \\ &=x^3+9x^2+27x+27\end{aligned} ( x + 3 ) 3 = ( x + 3 ) ( x + 3 ) 2 = ( x + 3 ) ( x 2 + 6 x + 9 ) = x ( x 2 + 6 x + 9 ) + 3 ( x 2 + 6 x + 9 ) = x 3 + 6 x 2 + 9 x + 3 x 2 + 1 8 x + 2 7 = x 3 + 9 x 2 + 2 7 x + 2 7
उदाहरण 2: ( 3 x + 4 y ) 3 = ( 3 x + 4 y ) ( 3 x + 4 y ) 2 = ( 3 x + 4 y ) ( 9 x 2 + 24 x y + 16 y 2 ) = 3 x ( 9 x 2 + 24 x y + 16 y 2 ) + 4 y ( 9 x 2 + 24 x y + 16 y 2 ) = 27 x 3 + 72 x 2 y + 48 x y 2 + 36 x 2 y + 96 x y 2 + 64 y 3 = 27 x 3 + 108 x 2 y + 144 x y 2 + 64 y 3 \begin{aligned} \\ \left(3x+4y\right)^3&=\left(3x+4y\right)\left(3x+4y\right)^2 \\ &=\left(3x+4y\right)\left(9x^2+24xy+16y^2\right) \\ &=3x\left(9x^2+24xy+16y^2 \right)+4y\left(9x^2+24xy+16y^2\right) \\ &=27x^3+72x^2 y+48xy^2+36x^2 y+96xy^2+64y^3 \\ &=27x^3+108x^2 y+144xy^2+64y^3\end{aligned} ( 3 x + 4 y ) 3 = ( 3 x + 4 y ) ( 3 x + 4 y ) 2 = ( 3 x + 4 y ) ( 9 x 2 + 2 4 x y + 1 6 y 2 ) = 3 x ( 9 x 2 + 2 4 x y + 1 6 y 2 ) + 4 y ( 9 x 2 + 2 4 x y + 1 6 y 2 ) = 2 7 x 3 + 7 2 x 2 y + 4 8 x y 2 + 3 6 x 2 y + 9 6 x y 2 + 6 4 y 3 = 2 7 x 3 + 1 0 8 x 2 y + 1 4 4 x y 2 + 6 4 y 3
उदाहरण 3: ( 2 m n + n 2 m ) 3 = 8 m 3 n 3 + 3 ( 4 m 2 n 2 n 2 m ) + 3 ( 2 m n 2 4 m 2 n ) + ( n 3 8 m 3 ) = 8 m 3 n 3 + 3 ( 4 m 2 n 2 n 2 m ) + 3 ( 2 m n 2 4 m 2 n ) + ( n 3 8 m 3 ) = 8 m 3 n 3 + 3 ( 2 m n ) + 3 ( n 2 m ) + ( n 3 8 m 3 ) = 8 m 3 n 3 + 6 m n + 3 n 2 m + n 3 8 m 3 \begin{aligned} \\ \left(\frac{2m}{n}+\frac{n}{2m}\right)^3&=\frac{8m^3}{n^3}+3\left(\frac{4m^2n}{2n^2m}\right)+3\left(\frac{2mn^2}{4m^2n}\right)+\left(\frac{n^3}{8m^3}\right) \\ &=\frac{8m^3}{n^3}+3\left(\frac{4m^2n}{2n^2m}\right)+3\left(\frac{2mn^2}{4m^2n}\right)+\left(\frac{n^3}{8m^3}\right) \\ &=\frac{8m^3}{n^3}+3\left(\frac{2m}{n}\right)+3\left(\frac{n}{2m}\right)+\left(\frac{n^3}{8m^3}\right) \\ &=\frac{8m^3}{n^3}+\frac{6m}{n}+\frac{3n}{2m}+\frac{n^3}{8m^3}\end{aligned} ( n 2 m + 2 m n ) 3 = n 3 8 m 3 + 3 ( 2 n 2 m 4 m 2 n ) + 3 ( 4 m 2 n 2 m n 2 ) + ( 8 m 3 n 3 ) = n 3 8 m 3 + 3 ( 2 n 2 m 4 m 2 n ) + 3 ( 4 m 2 n 2 m n 2 ) + ( 8 m 3 n 3 ) = n 3 8 m 3 + 3 ( n 2 m ) + 3 ( 2 m n ) + ( 8 m 3 n 3 ) = n 3 8 m 3 + n 6 m + 2 m 3 n + 8 m 3 n 3
उदाहरण 4: ( 41 ) 3 = ( 40 + 1 ) 3 = 4 0 3 + 3 ( 4 0 2 × 1 ) + 3 ( 40 × 1 2 ) + 1 3 = 64000 + 4800 + 120 + 1 = 68921 \begin{aligned} \\ \left(41\right)^3&=\left(40+1\right)^3\\&=40^3+3\left(40^2\times 1\right)+3\left(40\times 1^2\right)+1^3 \\ &=64000+4800+120+1\\&=68921\end{aligned} ( 4 1 ) 3 = ( 4 0 + 1 ) 3 = 4 0 3 + 3 ( 4 0 2 × 1 ) + 3 ( 4 0 × 1 2 ) + 1 3 = 6 4 0 0 0 + 4 8 0 0 + 1 2 0 + 1 = 6 8 9 2 1
आता आपण ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 \mathbf{\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3} ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3
उदाहरण 1: ( k + 4 ) 3 \left(k+4\right)^3 ( k + 4 ) 3 उत्तर: a = k , b = 4 a=k,b=4 a = k , b = 4 ∴ ( k + 4 ) 3 = k 3 + 3 ( k 2 4 ) + 3 ( k 4 2 ) + 4 3 = k 3 + 12 k 2 + 48 k + 64 \begin{aligned} \\ \therefore \left(k+4\right)^3&=k^3+3\left(k^24\right)+3\left(k4^2\right)+4^3 \\ &=k^3+12k^2+48k+64\end{aligned} ∴ ( k + 4 ) 3 = k 3 + 3 ( k 2 4 ) + 3 ( k 4 2 ) + 4 3 = k 3 + 1 2 k 2 + 4 8 k + 6 4
उदाहरण 2: ( 7 x + 8 y ) 3 \left(7x+8y\right)^3 ( 7 x + 8 y ) 3 उत्तर: a = 7 x , y = 8 y a=7x,y=8y a = 7 x , y = 8 y ∴ ( 7 x + 8 y ) 3 = ( 7 x ) 3 + [ 3 × ( 7 x ) 2 × 8 y ] + [ 3 × 7 x × ( 8 y ) 2 ] + ( 8 y ) 3 = 343 x 3 + 1176 x 2 y + 1344 x y 2 + 512 y 3 \begin{aligned}\therefore \left(7x+8y\right)^3&=\left(7x\right)^3+[3\times \left(7x\right)^2\times 8y]+[3\times 7x\times \left(8y\right)^2]+\left(8y\right)^3 \\ &=343x^3+1176x^2y+1344xy^2+512y^3\end{aligned} ∴ ( 7 x + 8 y ) 3 = ( 7 x ) 3 + [ 3 × ( 7 x ) 2 × 8 y ] + [ 3 × 7 x × ( 8 y ) 2 ] + ( 8 y ) 3 = 3 4 3 x 3 + 1 1 7 6 x 2 y + 1 3 4 4 x y 2 + 5 1 2 y 3
उदाहरण 3: ( 7 + m ) 3 \left(7+m\right)^3 ( 7 + m ) 3 उत्तर: a = 7 , b = m a=7,b=m a = 7 , b = m ∴ ( 7 + m ) 3 = 7 3 + 3 ( 7 2 × m ) + 3 ( 7 × m 2 ) + m 3 = 343 + 147 m + 21 m 2 + m 3 \begin{aligned} \\ \therefore \left(7+m\right)^3&=7^3+3\left(7^2\times m\right)+3\left(7\times m^2\right)+m^3 \\ &=343+147m+21m^2+m^3\end{aligned} ∴ ( 7 + m ) 3 = 7 3 + 3 ( 7 2 × m ) + 3 ( 7 × m 2 ) + m 3 = 3 4 3 + 1 4 7 m + 2 1 m 2 + m 3
उदाहरण 4: ( 52 ) 3 = ( 50 + 2 ) 3 \left(52\right)^3=\left(50+2\right)^3 ( 5 2 ) 3 = ( 5 0 + 2 ) 3 उत्तर: a = 50 , b = 2 a=50,b=2 a = 5 0 , b = 2 ∴ ( 50 + 2 ) 3 = 5 0 3 + 3 ( 5 0 2 × 2 ) + 3 ( 50 × 2 2 ) + 2 3 = 125000 + 15000 + 600 + 8 = 140608 \begin{aligned} \\ \therefore \left(50+2\right)^3&=50^3+3\left(50^2\times2\right)+3\left(50\times2^2\right)+2^3 \\ &=125000+15000+600+8 \\ &=140608\end{aligned} ∴ ( 5 0 + 2 ) 3 = 5 0 3 + 3 ( 5 0 2 × 2 ) + 3 ( 5 0 × 2 2 ) + 2 3 = 1 2 5 0 0 0 + 1 5 0 0 0 + 6 0 0 + 8 = 1 4 0 6 0 8
उदाहरण 5: ( 101 ) 3 = ( 100 + 1 ) 3 \left(101\right)^3=\left(100+1\right)^3 ( 1 0 1 ) 3 = ( 1 0 0 + 1 ) 3 उत्तर: a = 100 , b = 1 a=100,b=1 a = 1 0 0 , b = 1 ∴ ( 100 + 1 ) 3 = 10 0 3 + 3 ( 10 0 2 × 1 ) + 3 ( 100 × 1 2 ) + 1 3 = 1000000 + 30000 + 300 + 1 = 1030301 \begin{aligned} \\ \therefore \left(100+1\right)^3&=100^3+3\left(100^2\times1\right)+3\left(100\times1^2\right)+1^3 \\ &=1000000+30000+300+1 \\ &=1030301\end{aligned} ∴ ( 1 0 0 + 1 ) 3 = 1 0 0 3 + 3 ( 1 0 0 2 × 1 ) + 3 ( 1 0 0 × 1 2 ) + 1 3 = 1 0 0 0 0 0 0 + 3 0 0 0 0 + 3 0 0 + 1 = 1 0 3 0 3 0 1
उदाहरण 6: ( x + 1 x ) 3 \left(x+\frac{1}{x}\right)^3 ( x + x 1 ) 3 उत्तर: a = x , b = 1 x a=x,b=\frac{1}{x} a = x , b = x 1 ∴ ( x + 1 x ) 3 = x 3 + 3 ( x 2 x ) + 3 ( x x 2 ) + ( 1 x ) 3 = x 3 + 3 x + 3 x + 1 x 3 \begin{aligned} \\ \therefore \left(x+\frac{1}{x}\right)^3&=x^3+3\left(\frac{x^2}{x}\right)+3\left(\frac{x}{x^2}\right)+\left(\frac{1}{x}\right)^3 \\ &=x^3+3x+\frac{3}{x}+\frac{1}{x^3}\end{aligned} ∴ ( x + x 1 ) 3 = x 3 + 3 ( x x 2 ) + 3 ( x 2 x ) + ( x 1 ) 3 = x 3 + 3 x + x 3 + x 3 1
उदाहरण 7: ( 2 m + 1 5 ) 3 \left(2m+\frac{1}{5}\right)^3 ( 2 m + 5 1 ) 3 उत्तर: a = 2 m , b = 1 5 a=2m,b=\frac{1}{5} a = 2 m , b = 5 1 ∴ ( 2 m + 1 5 ) 3 = 8 m 3 + 3 ( 4 m 2 5 ) + 3 ( 2 m 5 2 ) + ( 1 5 ) 3 = 8 m 3 + ( 12 m 2 5 ) + ( 6 m 25 ) + ( 1 125 ) = 8 m 3 + 12 m 2 5 + 6 m 25 + 1 125 \begin{aligned} \\ \therefore \left(2m+\frac{1}{5}\right)^3&=8m^3+3\left(\frac{4m^2}{5}\right)+3\left(\frac{2m}{5^2}\right)+\left(\frac{1}{5}\right)^3 \\ &=8m^3+\left(\frac{12m^2}{5}\right)+\left(\frac{6m}{25}\right)+\left(\frac{1}{125}\right) \\ &=8m^3+\frac{12m^2}{5}+\frac{6m}{25}+\frac{1}{125}\end{aligned} ∴ ( 2 m + 5 1 ) 3 = 8 m 3 + 3 ( 5 4 m 2 ) + 3 ( 5 2 2 m ) + ( 5 1 ) 3 = 8 m 3 + ( 5 1 2 m 2 ) + ( 2 5 6 m ) + ( 1 2 5 1 ) = 8 m 3 + 5 1 2 m 2 + 2 5 6 m + 1 2 5 1
उदाहरण 8: ( 5 x y + y 5 x ) 3 \left(\frac{5x}{y}+\frac{y}{5x}\right)^3 ( y 5 x + 5 x y ) 3 उत्तर: a = 5 x y , b = y 5 x a=\frac{5x}{y},b=\frac{y}{5x} a = y 5 x , b = 5 x y ∴ ( 5 x y + y 5 x ) 3 = ( 5 x y ) 3 + 3 ( 25 x 2 y 2 × y 5 x ) + 3 ( 5 x y × y 2 25 x 2 ) + ( y 5 x ) 3 = ( 5 x y ) 3 + 3 ( 5 x y ) + 3 ( y 5 x ) + ( y 5 x ) 3 = 125 x 3 y 3 + 15 x y + 3 y 5 x + y 3 125 x 3 \begin{aligned}\therefore \left(\frac{5x}{y}+\frac{y}{5x}\right)^3&=\left(\frac{5x}{y}\right)^3+3\left(\frac{25x^2}{y^2}\times \frac{y}{5x}\right)+3\left(\frac{5x}{y}\times \frac{y^2}{25x^2}\right)+\left(\frac{y}{5x}\right)^3 \\ &=\left(\frac{5x}{y}\right)^3+3\left(\frac{5x}{y}\right)+3\left(\frac{y}{5x}\right)+\left(\frac{y}{5x}\right)^3 \\ &=\frac{125x^3}{y^3}+\frac{15x}{y}+\frac{3y}{5x}+\frac{y^3}{125x^3}\end{aligned} ∴ ( y 5 x + 5 x y ) 3 = ( y 5 x ) 3 + 3 ( y 2 2 5 x 2 × 5 x y ) + 3 ( y 5 x × 2 5 x 2 y 2 ) + ( 5 x y ) 3 = ( y 5 x ) 3 + 3 ( y 5 x ) + 3 ( 5 x y ) + ( 5 x y ) 3 = y 3 1 2 5 x 3 + y 1 5 x + 5 x 3 y + 1 2 5 x 3 y 3
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( a − b ) 3 \mathbf{\left(a-b\right)^3} ( a − b ) 3 चा विस्तार ( a − b ) 3 = ( a − b ) ( a − b ) 2 = ( a − b ) ( a − b ) 2 = ( a − b ) ( a 2 − 2 a b + b 2 ) = a ( a 2 − 2 a b + b 2 ) − b ( a 2 − 2 a b + b 2 ) = a 3 − 2 a 2 b + a b 2 − a 2 b + 2 a b 2 − b 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 \begin{aligned}\left(a-b\right)^3&=\left(a-b\right)\left(a-b\right)^2 \\ &=\left(a-b\right)\left(a-b\right)^2 \\ &=\left(a-b\right)\left(a^2-2ab+b^2\right) \\ &=a\left(a^2-2ab+b^2\right)-b\left(a^2-2ab+b^2\right) \\ &=a^3-2a^2b+ab^2-a^2b+2ab^2-b^3 \\ &=a^3-3a^2b+3ab^2-b^3\end{aligned} ( a − b ) 3 = ( a − b ) ( a − b ) 2 = ( a − b ) ( a − b ) 2 = ( a − b ) ( a 2 − 2 a b + b 2 ) = a ( a 2 − 2 a b + b 2 ) − b ( a 2 − 2 a b + b 2 ) = a 3 − 2 a 2 b + a b 2 − a 2 b + 2 a b 2 − b 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3
( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 \begin{aligned}\mathbf{\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3}\end{aligned} ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3
( a − b ) 3 \left(a-b\right)^3 ( a − b ) 3
उदाहरण 1: ( x − 2 ) 3 \left(x-2\right)^3 ( x − 2 ) 3 उत्तर: a = x , b = 2 a=x,b=2 a = x , b = 2 ( x − 2 ) 3 = x 3 − 3 ( x 2 × 2 ) + 3 ( x × 2 2 ) − 2 3 = x 3 − 6 x 2 + 12 x − 8 \begin{aligned} \\ \left(x-2\right)^3&=x^3-3\left(x^2\times 2\right)+3\left(x\times 2^2\right)-2^3 \\ &=x^3-6x^2+12x-8\end{aligned} ( x − 2 ) 3 = x 3 − 3 ( x 2 × 2 ) + 3 ( x × 2 2 ) − 2 3 = x 3 − 6 x 2 + 1 2 x − 8
उदाहरण 2: ( 4 p − 5 q ) 3 \left(4p-5q\right)^3 ( 4 p − 5 q ) 3 उत्तर: a = 4 p , b = 5 q a=4p,b=5q a = 4 p , b = 5 q ( 4 p − 5 q ) 3 = ( 4 p ) 3 − 3 ( ( 4 p ) 2 × 5 q ) + 3 ( 4 p × ( 5 q ) 2 ) − ( 5 q ) 3 = 64 p 3 − 240 p 2 q + 300 p q 2 − 125 q 3 \begin{aligned} \\ \left(4p-5q\right)^3&=\left(4p\right)^3-3\left(\left(4p\right)^2\times 5q\right)+3\left(4p\times \left(5q\right)^2\right)-\left(5q\right)^3 \\ &=64p^3-240p^2q+300pq^2-125q^3\end{aligned} ( 4 p − 5 q ) 3 = ( 4 p ) 3 − 3 ( ( 4 p ) 2 × 5 q ) + 3 ( 4 p × ( 5 q ) 2 ) − ( 5 q ) 3 = 6 4 p 3 − 2 4 0 p 2 q + 3 0 0 p q 2 − 1 2 5 q 3
उदाहरण 3: ( 99 ) 3 = ( 100 − 1 ) 3 \left(99\right)^3=\left(100-1\right)^3 ( 9 9 ) 3 = ( 1 0 0 − 1 ) 3 उत्तर: a = 100 , b = 1 a=100,b=1 a = 1 0 0 , b = 1 ( 100 − 1 ) 3 = 10 0 3 − 3 ( 10 0 2 × 1 ) + 3 ( 100 × 1 2 ) − 1 3 = 1000000 − 30000 + 300 − 1 = 970299 \begin{aligned} \\ \left(100-1\right)^3&=100^3-3\left(100^2\times 1\right)+3\left(100\times 1^2\right)-1^3 \\ &=1000000-30000+300-1 \\ &=970299\end{aligned} ( 1 0 0 − 1 ) 3 = 1 0 0 3 − 3 ( 1 0 0 2 × 1 ) + 3 ( 1 0 0 × 1 2 ) − 1 3 = 1 0 0 0 0 0 0 − 3 0 0 0 0 + 3 0 0 − 1 = 9 7 0 2 9 9
उदाहरण 4: ( p + q ) 3 + ( p − q ) 3 \left(p+q\right)^3+\left(p-q\right)^3 ( p + q ) 3 + ( p − q ) 3 उत्तर: a = p , b = q a=p,b=q a = p , b = q ( p + q ) 3 \left(p+q\right)^3 ( p + q ) 3 ( a + b ) 3 \left(a+b\right)^3 ( a + b ) 3 ( p − q ) 3 \left(p-q\right)^3 ( p − q ) 3 ( a − b ) 3 \left(a-b\right)^3 ( a − b ) 3 ∴ ( p + q ) 3 + ( p − q ) 3 = [ p 3 + 3 p 2 q + 3 p q 2 + q 3 ] + [ p 3 − 3 p 2 q + 3 p q 2 − q 3 ] = p 3 + 3 p 2 q + 3 p q 2 + q 3 + p 3 − 3 p 2 q + 3 p q 2 − q 3 = 2 p 3 + 6 p q 2 \begin{aligned} \\ \therefore \left(p+q\right)^3+\left(p-q\right)^3=&[p^3+3p^2q+3pq^2+q^3]+ \\ &[p^3-3p^2q+3pq^2-q^3] \\ =&\;p^3+\cancel{3p^2q}+3pq^2+\cancel{q^3}+ \\ &\;p^3-\cancel{3p^2q}+3pq^2-\cancel{q^3} \\ =&\;2p^3+6pq^2\end{aligned} ∴ ( p + q ) 3 + ( p − q ) 3 = = = [ p 3 + 3 p 2 q + 3 p q 2 + q 3 ] + [ p 3 − 3 p 2 q + 3 p q 2 − q 3 ] p 3 + 3 p 2 q + 3 p q 2 + q 3 + p 3 − 3 p 2 q + 3 p q 2 − q 3 2 p 3 + 6 p q 2
उदाहरण 5: ( 2 x + 3 y ) 3 − ( 2 x − 3 y ) 3 \left(2x+3y\right)^3-\left(2x-3y\right)^3 ( 2 x + 3 y ) 3 − ( 2 x − 3 y ) 3 उत्तर: a = 2 x , b = 3 y a=2x,b=3y a = 2 x , b = 3 y ( 2 x + 3 y ) 3 \left(2x+3y\right)^3 ( 2 x + 3 y ) 3 ( a + b ) 3 \left(a+b\right)^3 ( a + b ) 3 ( 2 x − 3 y ) 3 \left(2x-3y\right)^3 ( 2 x − 3 y ) 3 ( a − b ) 3 \left(a-b\right)^3 ( a − b ) 3 ( 2 x + 3 y ) 3 − ( 2 x − 3 y ) 3 = ( 2 x ) 3 + 3 ( ( 2 x ) 2 × 3 y ) + 3 ( 2 x × ( 3 y ) 2 ) + ( 3 y ) 3 ] − [ ( 2 x ) 3 − 3 ( 2 x ) 2 × 3 y + 3 ( 2 x × ( 3 y ) 2 ) − ( 3 y ) 3 ] = [ 8 x 3 + 36 x 2 y + 54 x y 2 + 27 y 3 ] − [ 8 x 3 − 36 x 2 y + 54 x y 2 − 27 x 3 ] = 8 x 3 + 36 x 2 y + 54 x y 2 + 27 y 3 − 8 x 3 + 36 x 2 y − 54 x y 2 + 27 x 3 = 72 x 2 y + 54 x 3 \begin{aligned} \\ \left(2x+3y\right)^3-\left(2x-3y\right)^3=& \;\left(2x\right)^3+3\left(\left(2x\right)^2\times3y\right)+3\left(2x\times\left(3y\right)^2\right)+\left(3y\right)^3]- \\ & \;[\left(2x\right)^3-3\left(2x\right)^2\times3y+3\left(2x\times\left(3y\right)^2\right)-\left(3y\right)^3] \\ =& \;[8x^3+36x^2y+54xy^2+27y^3]-\\ & \;[8x^3-36x^2y+54xy^2-27x^3] \\ =& \;\cancel{8x^3}+36x^2y+\cancel{54xy^2}+27y^3- \\ & \;\cancel{8x^3}+36x^2y-\cancel{54xy^2}+27x^3 \\ =& \;72x^2y+54x^3\end{aligned} ( 2 x + 3 y ) 3 − ( 2 x − 3 y ) 3 = = = = ( 2 x ) 3 + 3 ( ( 2 x ) 2 × 3 y ) + 3 ( 2 x × ( 3 y ) 2 ) + ( 3 y ) 3 ] − [ ( 2 x ) 3 − 3 ( 2 x ) 2 × 3 y + 3 ( 2 x × ( 3 y ) 2 ) − ( 3 y ) 3 ] [ 8 x 3 + 3 6 x 2 y + 5 4 x y 2 + 2 7 y 3 ] − [ 8 x 3 − 3 6 x 2 y + 5 4 x y 2 − 2 7 x 3 ] 8 x 3 + 3 6 x 2 y + 5 4 x y 2 + 2 7 y 3 − 8 x 3 + 3 6 x 2 y − 5 4 x y 2 + 2 7 x 3 7 2 x 2 y + 5 4 x 3
आता आपण ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 \left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3 ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3
उदाहरण 1: ( 2 m − 5 ) 3 \left(2m-5\right)^3 ( 2 m − 5 ) 3 उत्तर: a = 2 m , b = 5 a=2m,b=5 a = 2 m , b = 5 ( 2 m − 5 ) 3 = ( 2 m ) 3 − 3 ( ( 2 m ) 2 × 5 ) + 3 ( 2 m × 5 2 ) − ( 5 ) 3 = 8 m 3 − 3 ( 20 m 2 ) + 3 ( 50 m ) − ( 125 ) = 8 m 3 − 60 m 2 + 150 m − 125 \begin{aligned} \\ \left(2m-5\right)^3&=\left(2m\right)^3-3\left(\left(2m\right)^2\times 5\right)+3\left(2m\times 5^2\right)-\left(5\right)^3 \\ &=8m^3-3\left(20m^2\right)+3\left(50m\right)-\left(125\right) \\ &=8m^3-60m^2+150m-125\end{aligned} ( 2 m − 5 ) 3 = ( 2 m ) 3 − 3 ( ( 2 m ) 2 × 5 ) + 3 ( 2 m × 5 2 ) − ( 5 ) 3 = 8 m 3 − 3 ( 2 0 m 2 ) + 3 ( 5 0 m ) − ( 1 2 5 ) = 8 m 3 − 6 0 m 2 + 1 5 0 m − 1 2 5
उदाहरण 2: ( 4 − p ) 3 \left(4-p\right)^3 ( 4 − p ) 3 उत्तर: a = 4 , b = p a=4,b=p a = 4 , b = p ( 4 − p ) 3 = ( 4 ) 3 − 3 ( 4 2 × p ) + 3 ( 4 × p 2 ) − ( p ) 3 = 64 − 48 p + 12 p 2 − p 3 \begin{aligned} \\ \left(4-p\right)^3&=\left(4\right)^3-3\left(4^2\times p\right)+3\left(4\times p^2\right)-\left(p\right)^3 \\ &=64-48p+12p^2-p^3\end{aligned} ( 4 − p ) 3 = ( 4 ) 3 − 3 ( 4 2 × p ) + 3 ( 4 × p 2 ) − ( p ) 3 = 6 4 − 4 8 p + 1 2 p 2 − p 3
उदाहरण 3: ( 7 x − 9 y ) 3 \left(7x-9y\right)^3 ( 7 x − 9 y ) 3 उत्तर: a = 7 x , b = 9 y a=7x,b=9y a = 7 x , b = 9 y ( 7 x − 9 y ) 3 = ( 7 x ) 3 − 3 ( ( 7 x ) 2 × 9 y ) + 3 ( 7 x × ( 9 y ) 2 ) − ( 9 y ) 3 = 343 x 3 − 1323 x 2 y + 1701 x y 2 − 729 y 3 \begin{aligned} \\ \left(7x-9y\right)^3=\left(7x\right)^3-3\left(\left(7x\right)^2\times9y\right)+3\left(7x\times\left(9y\right)^2\right)-\left(9y\right)^3 \\ &=343x^3-1323x^2y+1701xy^2-729y^3\end{aligned} ( 7 x − 9 y ) 3 = ( 7 x ) 3 − 3 ( ( 7 x ) 2 × 9 y ) + 3 ( 7 x × ( 9 y ) 2 ) − ( 9 y ) 3 = 3 4 3 x 3 − 1 3 2 3 x 2 y + 1 7 0 1 x y 2 − 7 2 9 y 3
उदाहरण 4: ( 58 ) 3 = ( 60 − 2 ) 3 \left(58\right)^3=\left(60-2\right)^3 ( 5 8 ) 3 = ( 6 0 − 2 ) 3 उत्तर: a = 60 , b = 2 a=60,b=2 a = 6 0 , b = 2 ( 60 − 2 ) 3 = ( 60 ) 3 − 3 ( ( 60 ) 2 × 2 ) + 3 ( 60 × ( 2 ) 2 ) − ( 2 ) 3 = 216000 − 21600 + 720 − 8 = 195112 \begin{aligned} \\ \left(60-2\right)^3&=\left(60\right)^3-3\left(\left(60\right)^2\times2\right)+3\left(60\times\left(2\right)^2\right)-\left(2\right)^3 \\ &=216000-21600+720-8 \\ &=195112\end{aligned} ( 6 0 − 2 ) 3 = ( 6 0 ) 3 − 3 ( ( 6 0 ) 2 × 2 ) + 3 ( 6 0 × ( 2 ) 2 ) − ( 2 ) 3 = 2 1 6 0 0 0 − 2 1 6 0 0 + 7 2 0 − 8 = 1 9 5 1 1 2
उदाहरण 5: ( 198 ) 3 = ( 200 − 2 ) 3 \left(198\right)^3=\left(200-2\right)^3 ( 1 9 8 ) 3 = ( 2 0 0 − 2 ) 3 उत्तर: a = 200 , b = 2 a=200,b=2 a = 2 0 0 , b = 2 ( 200 − 2 ) 3 = ( 200 ) 3 − 3 ( ( 200 ) 2 × 2 ) + 3 ( 200 × ( 2 ) 2 ) − ( 2 ) 3 = 8000000 − 240000 + 2400 − 8 = 7762392 \begin{aligned} \\ \left(200-2\right)^3&=\left(200\right)^3-3\left(\left(200\right)^2\times2\right)+3\left(200\times\left(2\right)^2\right)-\left(2\right)^3 \\ &=8000000-240000+2400-8 \\ &=7762392\end{aligned} ( 2 0 0 − 2 ) 3 = ( 2 0 0 ) 3 − 3 ( ( 2 0 0 ) 2 × 2 ) + 3 ( 2 0 0 × ( 2 ) 2 ) − ( 2 ) 3 = 8 0 0 0 0 0 0 − 2 4 0 0 0 0 + 2 4 0 0 − 8 = 7 7 6 2 3 9 2
उदाहरण 6: ( 2 p − 1 2 p ) 3 \left(2p-\frac{1}{2p}\right)^3 ( 2 p − 2 p 1 ) 3 उत्तर: a = 2 p , b = 1 2 p a=2p,b=\frac{1}{2p} a = 2 p , b = 2 p 1 ( 2 p − 1 2 p ) 3 = ( 2 p ) 3 − 3 [ ( 2 p ) 2 × 1 2 p ] + 3 [ 2 p × ( 1 2 p ) 2 ] − ( 1 2 p ) 3 = 8 p 3 − 6 p + 3 2 p − 1 8 p 3 \begin{aligned} \\ \left(2p-\frac{1}{2p}\right)^3&=\left(2p\right)^3-3\left[\left(2p\right)^2\times\frac{1}{2p}\right]+3\left[2p\times\left(\frac{1}{2p}\right)^2\right]-\left(\frac{1}{2p}\right)^3 \\ &=8p^3-6p+\frac{3}{2p}-\frac{1}{8p^3}\end{aligned} ( 2 p − 2 p 1 ) 3 = ( 2 p ) 3 − 3 [ ( 2 p ) 2 × 2 p 1 ] + 3 [ 2 p × ( 2 p 1 ) 2 ] − ( 2 p 1 ) 3 = 8 p 3 − 6 p + 2 p 3 − 8 p 3 1
उदाहरण 7: ( 1 − 1 a ) 3 \left(1-\frac{1}{a}\right)^3 ( 1 − a 1 ) 3 उत्तर: a = 1 , b = 1 a a=1,b=\frac{1}{a} a = 1 , b = a 1 ( 1 − 1 a ) 3 = ( 1 ) 3 − 3 [ ( 1 ) 2 × 1 a ] + 3 [ 1 × ( 1 a ) 2 ] − ( 1 a ) 3 = 1 − 3 a + 3 a 2 − 1 a 3 \begin{aligned} \\ \left(1-\frac{1}{a}\right)^3&=\left(1\right)^3-3\left[\left(1\right)^2\times\frac{1}{a}\right]+3\left[1\times\left(\frac{1}{a}\right)^2\right]-\left(\frac{1}{a}\right)^3 \\ &=1-\frac{3}{a}+\frac{3}{a^2}-\frac{1}{a^3}\end{aligned} ( 1 − a 1 ) 3 = ( 1 ) 3 − 3 [ ( 1 ) 2 × a 1 ] + 3 [ 1 × ( a 1 ) 2 ] − ( a 1 ) 3 = 1 − a 3 + a 2 3 − a 3 1
उदाहरण 8: ( x 3 − 3 x ) 3 \left(\frac{x}{3}-\frac{3}{x}\right)^3 ( 3 x − x 3 ) 3 उत्तर: a = x 3 , b = 3 x a=\frac{x}{3},b=\frac{3}{x} a = 3 x , b = x 3 ( x 3 − 3 x ) 3 = ( x 3 ) 3 − 3 [ ( x 3 ) 2 × 3 x ] + 3 [ x 3 × ( 3 x ) 2 ] − ( 3 x ) 3 = x 3 27 − 3 ( 3 x 2 9 x ) + 3 ( 9 x 3 x 2 ) − 27 x 3 = x 3 27 − ( 9 x 2 9 x ) + ( 27 x 3 x 2 ) − 27 x 3 = x 3 27 − x + 9 x − 27 x 3 \begin{aligned} \\ \left(\frac{x}{3}-\frac{3}{x}\right)^3&=\left(\frac{x}{3}\right)^3-3\left[\left(\frac{x}{3}\right)^2\times \frac{3}{x}\right]+3\left[\frac{x}{3}\times \left(\frac{3}{x}\right)^2\right]-\left(\frac{3}{x}\right)^3 \\ &=\frac{x^3}{27}-3\left(\frac{3x^2}{9x}\right)+3\left(\frac{9x}{3x^2}\right)-\frac{27}{x^3} \\ &=\frac{x^3}{27}-\left(\frac{9x^2}{9x}\right)+\left(\frac{27x}{3x^2}\right)-\frac{27}{x^3} \\ &=\frac{x^3}{27}-x+\frac{9}{x}-\frac{27}{x^3}\end{aligned} ( 3 x − x 3 ) 3 = ( 3 x ) 3 − 3 [ ( 3 x ) 2 × x 3 ] + 3 [ 3 x × ( x 3 ) 2 ] − ( x 3 ) 3 = 2 7 x 3 − 3 ( 9 x 3 x 2 ) + 3 ( 3 x 2 9 x ) − x 3 2 7 = 2 7 x 3 − ( 9 x 9 x 2 ) + ( 3 x 2 2 7 x ) − x 3 2 7 = 2 7 x 3 − x + x 9 − x 3 2 7
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( a + b ) 3 \mathbf{\left(a+b\right)^3} ( a + b ) 3 आणि ( a − b ) 3 \mathbf{\left(a-b\right)^3} ( a − b ) 3 ची उदाहरणं खाली दिलेली ( a + b ) 3 \left(a+b\right)^3 ( a + b ) 3 ( a − b ) 3 \left(a-b\right)^3 ( a − b ) 3 ( a + b ) 3 \left(a+b\right)^3 ( a + b ) 3 ( a − b ) 3 \left(a-b\right)^3 ( a − b ) 3 ( a + b ) 3 \left(a+b\right)^3 ( a + b ) 3 ( a − b ) 3 \left(a-b\right)^3 ( a − b ) 3
( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 \mathbf{\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3} ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 \mathbf{\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3} ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3
चला तर मग आता ही दोन्ही विस्तार सूत्र वापरून अजून काही उदाहरणं सोडवूयात. इथे लक्षात घ्या की आपल्याला ( a + b ) 3 \left(a+b\right)^3 ( a + b ) 3 ( a − b ) 3 \left(a-b\right)^3 ( a − b ) 3
उदाहरण 1: ( 2 a + b ) 3 − ( 2 a − b ) 3 \left(2a+b\right)^3-\left(2a-b\right)^3 ( 2 a + b ) 3 − ( 2 a − b ) 3 उत्तर: ∴ ( 2 a + b ) 3 = ( 2 a ) 3 + 3 ( ( 2 a ) 2 × b ) + 3 ( 2 a × ( b ) 2 ) + ( b ) 3 = 8 a 3 + 3 ( 4 a 2 b ) + 3 ( 2 a b 2 ) + b 3 \begin{aligned} \\ \therefore\left(2a+b\right)^3=&\left(2a\right)^3+3\left(\left(2a\right)^2\times b\right)+ \\ &\;3\left(2a\times \left(b\right)^2\right)+\left(b\right)^3 \\ =&\;8a^3+3\left(4a^2b\right)+3\left(2ab^2\right)+b^3\end{aligned} ∴ ( 2 a + b ) 3 = = ( 2 a ) 3 + 3 ( ( 2 a ) 2 × b ) + 3 ( 2 a × ( b ) 2 ) + ( b ) 3 8 a 3 + 3 ( 4 a 2 b ) + 3 ( 2 a b 2 ) + b 3 ∴ ( 2 a − b ) 3 = ( 2 a ) 3 − 3 ( ( 2 a ) 2 × b ) + 3 ( 2 a × ( b ) 2 ) − ( b ) 3 = 8 a 3 − 3 ( 4 a 2 b ) + 3 ( 2 a b 2 ) − b 3 \begin{aligned} \\ \therefore\left(2a-b\right)^3=&\left(2a\right)^3-3\left(\left(2a\right)^2\times b\right)+ \\ &\;3\left(2a\times \left(b\right)^2\right)-\left(b\right)^3 \\ =&\;8a^3-3\left(4a^2b\right)+3\left(2ab^2\right)-b^3\end{aligned} ∴ ( 2 a − b ) 3 = = ( 2 a ) 3 − 3 ( ( 2 a ) 2 × b ) + 3 ( 2 a × ( b ) 2 ) − ( b ) 3 8 a 3 − 3 ( 4 a 2 b ) + 3 ( 2 a b 2 ) − b 3 ∴ ( 2 a + b ) 3 − ( 2 a − b ) 3 = 8 a 3 + 12 a 2 b + 6 a b 2 + b 3 − 8 a 3 + 12 a 2 b − 6 a b 2 + b 3 = 24 a 2 b + 2 b 3 \begin{aligned} \\ \therefore\left(2a+b\right)^3-\left(2a-b\right)^3=&\;\cancel{8a^3}+12a^2b+\cancel{6ab^2}+b^3- \\ &\;\cancel{8a^3}+12a^2b-\cancel{6ab^2}+b^3 \\ =&\;24a^2b+2b^3\end{aligned} ∴ ( 2 a + b ) 3 − ( 2 a − b ) 3 = = 8 a 3 + 1 2 a 2 b + 6 a b 2 + b 3 − 8 a 3 + 1 2 a 2 b − 6 a b 2 + b 3 2 4 a 2 b + 2 b 3
उदाहरण 2: ( 3 r − 2 k ) 3 + ( 3 r + 2 k ) 3 \left(3r-2k\right)^3+\left(3r+2k\right)^3 ( 3 r − 2 k ) 3 + ( 3 r + 2 k ) 3 उत्तर: ∴ ( 3 r − 2 k ) 3 = ( 3 r ) 3 − 3 ( ( 3 r ) 2 × 2 k ) + 3 ( 3 r × ( 2 k ) 2 ) − ( 2 k ) 3 = 27 r 3 − 3 ( 18 r 2 k ) + 3 ( 12 r k 2 ) − 8 k 3 \begin{aligned} \\ \therefore\left(3r-2k\right)^3=&\left(3r\right)^3-3\left(\left(3r\right)^2\times2k\right)+ \\ &\;3\left(3r\times\left(2k\right)^2\right)-\left(2k\right)^3 \\ =&\;27r^3-3\left(18r^2k\right)+3\left(12rk^2\right)-8k^3\end{aligned} ∴ ( 3 r − 2 k ) 3 = = ( 3 r ) 3 − 3 ( ( 3 r ) 2 × 2 k ) + 3 ( 3 r × ( 2 k ) 2 ) − ( 2 k ) 3 2 7 r 3 − 3 ( 1 8 r 2 k ) + 3 ( 1 2 r k 2 ) − 8 k 3 ∴ ( 3 r + 2 k ) 3 = ( 3 r ) 3 + 3 ( ( 3 r ) 2 × 2 k ) + 3 ( 3 r × ( 2 k ) 2 ) + ( 2 k ) 3 = 27 r 3 + 3 ( 18 r 2 k ) + 3 ( 12 r k 2 ) + 8 k 3 \begin{aligned} \\ \therefore\left(3r+2k\right)^3=&\left(3r\right)^3+3\left(\left(3r\right)^2\times2k\right)+ \\ &\;3\left(3r\times\left(2k\right)^2\right)+\left(2k\right)^3 \\ =&\;27r^3+3\left(18r^2k\right)+3\left(12rk^2\right)+8k^3\end{aligned} ∴ ( 3 r + 2 k ) 3 = = ( 3 r ) 3 + 3 ( ( 3 r ) 2 × 2 k ) + 3 ( 3 r × ( 2 k ) 2 ) + ( 2 k ) 3 2 7 r 3 + 3 ( 1 8 r 2 k ) + 3 ( 1 2 r k 2 ) + 8 k 3 ( 3 r − 2 k ) 3 + ( 3 r + 2 k ) 3 = 27 r 3 − 54 r 2 k + 36 r k 2 − 8 k 3 + 27 r 3 + 54 r 2 k + 36 r k 2 + 8 k 3 = 54 r 3 + 72 r k 2 \begin{aligned} \\ \left(3r-2k\right)^3+\left(3r+2k\right)^3=&\;27r^3-\cancel{54r^2k}+36rk^2-\cancel{8k^3}+ \\ &\;27r^3+\cancel{54r^2k}+36rk^2+\cancel{8k^3} \\ =&\;54r^3+72rk^2\end{aligned} ( 3 r − 2 k ) 3 + ( 3 r + 2 k ) 3 = = 2 7 r 3 − 5 4 r 2 k + 3 6 r k 2 − 8 k 3 + 2 7 r 3 + 5 4 r 2 k + 3 6 r k 2 + 8 k 3 5 4 r 3 + 7 2 r k 2
उदाहरण 3: ( 4 a − 3 ) 3 − ( 4 a + 3 ) 3 \left(4a-3\right)^3-\left(4a+3\right)^3 ( 4 a − 3 ) 3 − ( 4 a + 3 ) 3 उत्तर: ∴ ( 4 a − 3 ) 3 = ( 4 a ) 3 − 3 ( ( 4 a ) 2 × 3 ) + 3 ( 4 a × ( 3 ) 2 ) − ( 3 ) 3 = 64 a 3 − 144 a 2 + 108 a − 27 \begin{aligned} \\ \therefore\left(4a-3\right)^3=&\left(4a\right)^3-3\left(\left(4a\right)^2\times 3\right)+ \\ &\;3\left(4a\times \left(3\right)^2\right)-\left(3\right)^3 \\ =&\;64a^3-144a^2+108a-27\end{aligned} ∴ ( 4 a − 3 ) 3 = = ( 4 a ) 3 − 3 ( ( 4 a ) 2 × 3 ) + 3 ( 4 a × ( 3 ) 2 ) − ( 3 ) 3 6 4 a 3 − 1 4 4 a 2 + 1 0 8 a − 2 7 ∴ ( 4 a + 3 ) 3 = ( 4 a ) 3 + 3 ( ( 4 a ) 2 × 3 ) + 3 ( 4 a × ( 3 ) 2 ) + ( 3 ) 3 = 64 a 3 − 144 a 2 − 108 a − 27 \begin{aligned} \\ \therefore\left(4a+3\right)^3=&\left(4a\right)^3+3\left(\left(4a\right)^2\times 3\right)+ \\ &\;3\left(4a\times \left(3\right)^2\right)+\left(3\right)^3 \\ =&\;64a^3-144a^2-108a-27\end{aligned} ∴ ( 4 a + 3 ) 3 = = ( 4 a ) 3 + 3 ( ( 4 a ) 2 × 3 ) + 3 ( 4 a × ( 3 ) 2 ) + ( 3 ) 3 6 4 a 3 − 1 4 4 a 2 − 1 0 8 a − 2 7 ( 4 a − 3 ) 3 − ( 4 a + 3 ) 3 = 64 a 3 − 144 a 2 + 108 a − 27 − 64 a 3 − 144 a 2 − 108 a − 27 = − 288 a 2 − 54 \begin{aligned} \\ \left(4a-3\right)^3-\left(4a+3\right)^3=&\;\cancel{64a^3}-144a^2+\cancel{108a}-27- \\ &\;\cancel{64a^3}-144a^2-\cancel{108a}-27 \\ =&\;-288a^2-54\end{aligned} ( 4 a − 3 ) 3 − ( 4 a + 3 ) 3 = = 6 4 a 3 − 1 4 4 a 2 + 1 0 8 a − 2 7 − 6 4 a 3 − 1 4 4 a 2 − 1 0 8 a − 2 7 − 2 8 8 a 2 − 5 4
उदाहरण 4: ( 5 x − 7 y ) 3 + ( 5 x + 7 y ) 3 \left(5x-7y\right)^3+\left(5x+7y\right)^3 ( 5 x − 7 y ) 3 + ( 5 x + 7 y ) 3 उत्तर: ∴ ( 5 x − 7 y ) 3 = ( 5 x ) 3 − 3 ( ( 5 x ) 2 × 7 y ) + 3 ( 5 x × ( 7 y ) 2 ) − ( 7 y ) 3 = 125 x 3 − 525 x 2 y + 735 x y 2 − 343 y 3 \begin{aligned} \\ \therefore\left(5x-7y\right)^3=&\left(5x\right)^3-3\left(\left(5x\right)^2\times 7y\right)+ \\ &\;3\left(5x\times \left(7y\right)^2\right)-\left(7y\right)^3 \\ =&\;125x^3-525x^2y+735xy^2-343y^3\end{aligned} ∴ ( 5 x − 7 y ) 3 = = ( 5 x ) 3 − 3 ( ( 5 x ) 2 × 7 y ) + 3 ( 5 x × ( 7 y ) 2 ) − ( 7 y ) 3 1 2 5 x 3 − 5 2 5 x 2 y + 7 3 5 x y 2 − 3 4 3 y 3 ∴ ( 5 x + 7 y ) 3 = ( 5 x ) 3 + 3 ( ( 5 x ) 2 × 7 y ) + 3 ( 5 x × ( 7 y ) 2 ) + ( 7 y ) 3 = 125 x 3 + 525 x 2 y + 735 x y 2 + 343 y 3 \begin{aligned} \\ \therefore\left(5x+7y\right)^3=&\left(5x\right)^3+3\left(\left(5x\right)^2\times 7y\right)+ \\ &\;3\left(5x\times \left(7y\right)^2\right)+\left(7y\right)^3 \\ =&\;125x^3+525x^2y+735xy^2+343y^3\end{aligned} ∴ ( 5 x + 7 y ) 3 = = ( 5 x ) 3 + 3 ( ( 5 x ) 2 × 7 y ) + 3 ( 5 x × ( 7 y ) 2 ) + ( 7 y ) 3 1 2 5 x 3 + 5 2 5 x 2 y + 7 3 5 x y 2 + 3 4 3 y 3 ∴ ( 5 x − 7 y ) 3 + ( 5 x + 7 y ) 3 = 125 x 3 − 525 x 2 y + 735 x y 2 − 343 y 3 + 125 x 3 + 525 x 2 y + 735 x y 2 + 343 y 3 = 250 x 3 + 1470 x y 2 \begin{aligned} \\ \therefore\left(5x-7y\right)^3+\left(5x+7y\right)^3=&\;125x^3-\cancel{525x^2y}+735xy^2-\cancel{343y^3}+ \\ &\;125x^3+\cancel{525x^2y}+735xy^2+\cancel{343y^3} \\ =&\;250x^3+1470xy^2\end{aligned} ∴ ( 5 x − 7 y ) 3 + ( 5 x + 7 y ) 3 = = 1 2 5 x 3 − 5 2 5 x 2 y + 7 3 5 x y 2 − 3 4 3 y 3 + 1 2 5 x 3 + 5 2 5 x 2 y + 7 3 5 x y 2 + 3 4 3 y 3 2 5 0 x 3 + 1 4 7 0 x y 2
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( a + b + c ) 2 \mathbf{\left(a+b+c\right)^2} ( a + b + c ) 2 ( a + b + c ) 2 = ( a + b + c ) × ( a + b + c ) = a ( a + b + c ) + b ( a + b + c ) + c ( a + b + c ) = a 2 + a b + a c + a b + b 2 + b c + a c + b c + c 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 a c \begin{aligned} \\ \left(a+b+c\right)^2&=\left(a+b+c\right)\times \left(a+b+c\right) \\ &=a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right) \\ &=a^2+ab+ac+ab+b^2+bc+ac+bc+c^2 \\ &=a^2+b^2+c^2+2ab+2bc+2ac\end{aligned} ( a + b + c ) 2 = ( a + b + c ) × ( a + b + c ) = a ( a + b + c ) + b ( a + b + c ) + c ( a + b + c ) = a 2 + a b + a c + a b + b 2 + b c + a c + b c + c 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 a c
( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 a c \mathbf{\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac} ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 a c
चला तर मग आता आपण ह्याचं विस्तार सूत्र वापरून काही उदाहरणं सोडवूयात,
उदाहरण 1: ( x + y + 3 ) 2 \left(x+y+3\right)^2 ( x + y + 3 ) 2 उत्तर: ∴ ( x + y + 3 ) 2 = x 2 + y 2 + 9 + 2 x y + 6 y + 6 x = x 2 + y 2 + 6 x + 6 y + 2 x y + 9 \begin{aligned} \\ \therefore \left(x+y+3\right)^2&=x^2+y^2+9+2xy+6y+6x \\ &=x^2+y^2+6x+6y+2xy+9\end{aligned} ∴ ( x + y + 3 ) 2 = x 2 + y 2 + 9 + 2 x y + 6 y + 6 x = x 2 + y 2 + 6 x + 6 y + 2 x y + 9
उदाहरण 2: ( 2 p + 3 q + 4 r ) 2 \left(2p+3q+4r\right)^2 ( 2 p + 3 q + 4 r ) 2 उत्तर: ∴ ( 2 p + 3 q + 4 r ) 2 = ( 2 p ) 2 + ( 3 q ) 2 + ( 4 r ) 2 + 2 ( 2 p × 3 q ) + 2 ( 3 q × 4 r ) + 2 ( 2 p × 4 r ) = 4 p 2 + 9 q 2 + 16 r 2 + 12 p q + 24 q r + 16 p r \begin{aligned} \\ \therefore \left(2p+3q+4r\right)^2=&\left(2p\right)^2+\left(3q\right)^2+\left(4r\right)^2+ \\ &\;2\left(2p\times 3q\right)+2\left(3q\times 4r\right)+2\left(2p\times 4r\right) \\ =&\;4p^2+9q^2+16r^2+12pq+24qr+16pr\end{aligned} ∴ ( 2 p + 3 q + 4 r ) 2 = = ( 2 p ) 2 + ( 3 q ) 2 + ( 4 r ) 2 + 2 ( 2 p × 3 q ) + 2 ( 3 q × 4 r ) + 2 ( 2 p × 4 r ) 4 p 2 + 9 q 2 + 1 6 r 2 + 1 2 p q + 2 4 q r + 1 6 p r
उदाहरण 3: ( a + 2 b + c ) 2 + ( a − 2 b + c ) 2 \left(a+2b+c\right)^2+\left(a-2b+c\right)^2 ( a + 2 b + c ) 2 + ( a − 2 b + c ) 2 उत्तर: ∴ ( a + 2 b + c ) 2 = ( a ) 2 + ( 2 b ) 2 + ( c ) 2 + 2 ( a × 2 b ) + 2 ( 2 b × c ) + 2 ( a × c ) = a 2 + 4 b 2 + c 2 + 4 a b + 4 b c + 2 a c \begin{aligned} \\ \therefore\left(a+2b+c\right)^2=&\left(a\right)^2+\left(2b\right)^2+\left(c\right)^2+ \\ &\;2\left(a\times 2b\right)+2\left(2b\times c\right)+2\left(a\times c\right) \\ =&\;a^2+4b^2+c^2+4ab+4bc+2ac\end{aligned} ∴ ( a + 2 b + c ) 2 = = ( a ) 2 + ( 2 b ) 2 + ( c ) 2 + 2 ( a × 2 b ) + 2 ( 2 b × c ) + 2 ( a × c ) a 2 + 4 b 2 + c 2 + 4 a b + 4 b c + 2 a c ∴ ( a − 2 b + c ) 2 = ( a ) 2 + ( 2 b ) 2 + ( c ) 2 − 2 ( a × 2 b ) − 2 ( 2 b × c ) + 2 ( a × c ) = a 2 + 4 b 2 + c 2 − 4 a b − 4 b c + 2 a c \begin{aligned} \\ \therefore\left(a-2b+c\right)^2=&\left(a\right)^2+\left(2b\right)^2+\left(c\right)^2- \\ &\;2\left(a\times 2b\right)-2\left(2b\times c\right)+2\left(a\times c\right) \\ =&\;a^2+4b^2+c^2-4ab-4bc+2ac\end{aligned} ∴ ( a − 2 b + c ) 2 = = ( a ) 2 + ( 2 b ) 2 + ( c ) 2 − 2 ( a × 2 b ) − 2 ( 2 b × c ) + 2 ( a × c ) a 2 + 4 b 2 + c 2 − 4 a b − 4 b c + 2 a c ∴ ( a + 2 b + c ) 2 + ( a − 2 b + c ) 2 = a 2 + 4 b 2 + c 2 + 4 a b + 4 b c + 2 a c + a 2 + 4 b 2 + c 2 − 4 a b − 4 b c + 2 a c = 2 a 2 + 8 b 2 + 2 c 2 + 4 a c \begin{aligned} \\ \therefore\left(a+2b+c\right)^2+\left(a-2b+c\right)^2=&\;a^2+4b^2+c^2+\cancel{4ab}+\cancel{4bc}+2ac+ \\ &\;a^2+4b^2+c^2-\cancel{4ab}-\cancel{4bc}+2ac \\ =&\;2a^2+8b^2+2c^2+4ac\end{aligned} ∴ ( a + 2 b + c ) 2 + ( a − 2 b + c ) 2 = = a 2 + 4 b 2 + c 2 + 4 a b + 4 b c + 2 a c + a 2 + 4 b 2 + c 2 − 4 a b − 4 b c + 2 a c 2 a 2 + 8 b 2 + 2 c 2 + 4 a c
उदाहरण 4: ( 2 m + n + 5 ) 2 \left(2m+n+5\right)^2 ( 2 m + n + 5 ) 2 उत्तर: ∴ ( 2 m + n + 5 ) 2 = ( 2 m ) 2 + ( n ) 2 + ( 5 ) 2 + 2 ( 2 m × n ) + 2 ( n × 5 ) + 2 ( 2 m × 5 ) = 4 m 2 + n 2 + 25 + 4 m n + 10 n + 20 m \begin{aligned} \\ \therefore \left(2m+n+5\right)^2=&\left(2m\right)^2+\left(n\right)^2+\left(5\right)^2+2\left(2m\times n\right)+ \\ &\;2\left(n\times 5\right)+2\left(2m\times 5\right) \\ =&\;4m^2+n^2+25+4mn+10n+20m\end{aligned} ∴ ( 2 m + n + 5 ) 2 = = ( 2 m ) 2 + ( n ) 2 + ( 5 ) 2 + 2 ( 2 m × n ) + 2 ( n × 5 ) + 2 ( 2 m × 5 ) 4 m 2 + n 2 + 2 5 + 4 m n + 1 0 n + 2 0 m
उदाहरण 5: ( p + 2 q + 3 r ) 2 \left(p+2q+3r\right)^2 ( p + 2 q + 3 r ) 2 उत्तर: ∴ ( p + 2 q + 3 r ) 2 = ( p ) 2 + ( 2 q ) 2 + ( 3 r ) 2 + 2 ( p × 2 q ) + 2 ( 2 q × 3 r ) + 2 ( p × 3 r ) = p 2 + 4 q 2 + 9 r 2 + 4 p q + 12 q r + 6 p r \begin{aligned} \\ \therefore \left(p+2q+3r\right)^2=&\left(p\right)^2+\left(2q\right)^2+\left(3r\right)^2+2\left(p\times 2q\right)+ \\ &\;2\left(2q\times 3r\right)+2\left(p\times 3r\right) \\ =&\;p^2+4q^2+9r^2+4pq+12qr+6pr\end{aligned} ∴ ( p + 2 q + 3 r ) 2 = = ( p ) 2 + ( 2 q ) 2 + ( 3 r ) 2 + 2 ( p × 2 q ) + 2 ( 2 q × 3 r ) + 2 ( p × 3 r ) p 2 + 4 q 2 + 9 r 2 + 4 p q + 1 2 q r + 6 p r
उदाहरण 6: ( 3 a + 4 b − 5 k ) 2 \left(3a+4b-5k\right)^2 ( 3 a + 4 b − 5 k ) 2 उत्तर: ∴ ( 3 a + 4 b − 5 k ) 2 = ( 3 a ) 2 + ( 4 b ) 2 + ( − 5 k ) 2 + 2 ( 3 a × 4 b ) + 2 ( 4 b × ( − 5 k ) ) + 2 ( 3 a × ( − 5 k ) ) = 9 a 2 + 16 b 2 + 25 k 2 + 24 a b − 40 b k − 30 a k \begin{aligned} \\ \therefore \left(3a+4b-5k\right)^2=&\left(3a\right)^2+\left(4b\right)^2+\left(-5k\right)^2+2\left(3a\times 4b\right)+ \\ &\;2\left(4b\times \left(-5k\right)\right)+2\left(3a\times \left(-5k\right)\right) \\ =&\;9a^2+16b^2+25k^2+24ab-40bk-30ak\end{aligned} ∴ ( 3 a + 4 b − 5 k ) 2 = = ( 3 a ) 2 + ( 4 b ) 2 + ( − 5 k ) 2 + 2 ( 3 a × 4 b ) + 2 ( 4 b × ( − 5 k ) ) + 2 ( 3 a × ( − 5 k ) ) 9 a 2 + 1 6 b 2 + 2 5 k 2 + 2 4 a b − 4 0 b k − 3 0 a k
उदाहरण 7: ( 7 p − 3 q − 4 r ) 2 \left(7p-3q-4r\right)^2 ( 7 p − 3 q − 4 r ) 2 उत्तर: ∴ ( 7 p − 3 q − 4 r ) 2 = ( 7 p ) 2 + ( − 3 q ) 2 + ( − 4 r ) 2 + 2 ( 7 p × ( − 3 q ) ) + 2 ( ( − 3 q ) × ( − 4 r ) ) + 2 ( 7 p × ( − 4 r ) ) = 49 p 2 + 9 q 2 + 16 r 2 − 42 p q + 24 q r − 56 p r \begin{aligned} \\ \therefore \left(7p-3q-4r\right)^2=&\left(7p\right)^2+\left(-3q\right)^2+\left(-4r\right)^2+2\left(7p\times \left(-3q\right)\right)+ \\ &\;2\left(\left(-3q\right)\times \left(-4r\right)\right)+2\left(7p\times \left(-4r\right)\right) \\ =&\;49p^2+9q^2+16r^2-42pq+24qr-56pr\end{aligned} ∴ ( 7 p − 3 q − 4 r ) 2 = = ( 7 p ) 2 + ( − 3 q ) 2 + ( − 4 r ) 2 + 2 ( 7 p × ( − 3 q ) ) + 2 ( ( − 3 q ) × ( − 4 r ) ) + 2 ( 7 p × ( − 4 r ) ) 4 9 p 2 + 9 q 2 + 1 6 r 2 − 4 2 p q + 2 4 q r − 5 6 p r
उदाहरण 8: ( a − 2 b + 3 ) 2 + ( a + 2 b − 3 ) 2 \left(a-2b+3\right)^2+\left(a+2b-3\right)^2 ( a − 2 b + 3 ) 2 + ( a + 2 b − 3 ) 2 उत्तर: ∴ ( a − 2 b + 3 ) 2 = ( a ) 2 + ( − 2 b ) 2 + ( 3 ) 2 + 2 ( a × ( − 2 b ) ) + 2 ( ( − 2 b ) × 3 ) + 2 ( a × 3 ) = a 2 + 4 b 2 + 9 − 4 a b − 12 b + 6 a \begin{aligned} \\ \therefore\left(a-2b+3\right)^2=&\left(a\right)^2+\left(-2b\right)^2+\left(3\right)^2+2\left(a\times \left(-2b\right)\right)+ \\ &\;2\left(\left(-2b\right)\times 3\right)+2\left(a\times 3\right) \\ =&\;a^2+4b^2+9-4ab-12b+6a\end{aligned} ∴ ( a − 2 b + 3 ) 2 = = ( a ) 2 + ( − 2 b ) 2 + ( 3 ) 2 + 2 ( a × ( − 2 b ) ) + 2 ( ( − 2 b ) × 3 ) + 2 ( a × 3 ) a 2 + 4 b 2 + 9 − 4 a b − 1 2 b + 6 a ∴ ( a + 2 b − 3 ) 2 = ( a ) 2 + ( 2 b ) 2 + ( − 3 ) 2 + 2 ( a × ( 2 b ) ) + 2 ( ( 2 b ) × ( − 3 ) ) + 2 ( a × ( − 3 ) ) = a 2 + 4 b 2 + 9 + 4 a b − 12 b − 6 a \begin{aligned} \\ \therefore\left(a+2b-3\right)^2=&\left(a\right)^2+\left(2b\right)^2+\left(-3\right)^2+2\left(a\times \left(2b\right)\right)+ \\ &\;2\left(\left(2b\right)\times \left(-3\right)\right)+2\left(a\times \left(-3\right)\right) \\ =&\;a^2+4b^2+9+4ab-12b-6a\end{aligned} ∴ ( a + 2 b − 3 ) 2 = = ( a ) 2 + ( 2 b ) 2 + ( − 3 ) 2 + 2 ( a × ( 2 b ) ) + 2 ( ( 2 b ) × ( − 3 ) ) + 2 ( a × ( − 3 ) ) a 2 + 4 b 2 + 9 + 4 a b − 1 2 b − 6 a ∴ ( a − 2 b + 3 ) 2 + ( a + 2 b − 3 ) 2 = a 2 + 4 b 2 + 9 − 4 a b − 12 b + 6 a + a 2 + 4 b 2 + 9 + 4 a b − 12 b − 6 a = 2 a 2 + 8 b 2 + 18 − 24 b \begin{aligned} \\ \therefore\left(a-2b+3\right)^2+\left(a+2b-3\right)^2=&\;a^2+4b^2+9-\cancel{4ab}-12b+\cancel{6a}+ \\ &\;a^2+4b^2+9+\cancel{4ab}-12b-\cancel{6a} \\ =&\;2a^2+8b^2+18-24b\end{aligned} ∴ ( a − 2 b + 3 ) 2 + ( a + 2 b − 3 ) 2 = = a 2 + 4 b 2 + 9 − 4 a b − 1 2 b + 6 a + a 2 + 4 b 2 + 9 + 4 a b − 1 2 b − 6 a 2 a 2 + 8 b 2 + 1 8 − 2 4 b
उदाहरण 9: ( 3 x − 4 y − 2 z ) 2 − ( 3 x + 4 y − 2 z ) 2 \left(3x-4y-2z\right)^2-\left(3x+4y-2z\right)^2 ( 3 x − 4 y − 2 z ) 2 − ( 3 x + 4 y − 2 z ) 2 उत्तर: ∴ ( 3 x − 4 y − 2 z ) 2 = ( 3 x ) 2 + ( − 4 y ) 2 + ( − 2 z ) 2 + 2 ( ( 3 x ) × ( − 4 y ) ) + 2 ( ( − 4 y ) × ( − 2 z ) ) + 2 ( ( 3 x ) × ( − 2 z ) ) = 9 x 2 + 16 y 2 + 4 z 2 − 24 x y + 16 y z − 12 x z \begin{aligned} \\ \therefore\left(3x-4y-2z\right)^2=&\left(3x\right)^2+\left(-4y\right)^2+\left(-2z\right)^2+2\left(\left(3x\right)\times \left(-4y\right)\right)+ \\ &\;2\left(\left(-4y\right)\times \left(-2z\right)\right)+2\left(\left(3x\right)\times \left(-2z\right)\right) \\ =&\;9x^2+16y^2+4z^2-24xy+16yz-12xz\end{aligned} ∴ ( 3 x − 4 y − 2 z ) 2 = = ( 3 x ) 2 + ( − 4 y ) 2 + ( − 2 z ) 2 + 2 ( ( 3 x ) × ( − 4 y ) ) + 2 ( ( − 4 y ) × ( − 2 z ) ) + 2 ( ( 3 x ) × ( − 2 z ) ) 9 x 2 + 1 6 y 2 + 4 z 2 − 2 4 x y + 1 6 y z − 1 2 x z ∴ ( 3 x + 4 y − 2 z ) 2 = ( 3 x ) 2 + ( 4 y ) 2 + ( − 2 z ) 2 + 2 ( ( 3 x ) × ( 4 y ) ) + 2 ( ( 4 y ) × ( − 2 z ) ) + 2 ( ( 3 x ) × ( − 2 z ) ) = 9 x 2 + 16 y 2 + 4 z 2 + 24 x y − 16 y z − 12 x z \begin{aligned} \\ \therefore\left(3x+4y-2z\right)^2=&\left(3x\right)^2+\left(4y\right)^2+\left(-2z\right)^2+2\left(\left(3x\right)\times \left(4y\right)\right)+ \\ &\;2\left(\left(4y\right)\times \left(-2z\right)\right)+2\left(\left(3x\right)\times \left(-2z\right)\right) \\ =&\;9x^2+16y^2+4z^2+24xy-16yz-12xz\end{aligned} ∴ ( 3 x + 4 y − 2 z ) 2 = = ( 3 x ) 2 + ( 4 y ) 2 + ( − 2 z ) 2 + 2 ( ( 3 x ) × ( 4 y ) ) + 2 ( ( 4 y ) × ( − 2 z ) ) + 2 ( ( 3 x ) × ( − 2 z ) ) 9 x 2 + 1 6 y 2 + 4 z 2 + 2 4 x y − 1 6 y z − 1 2 x z ∴ ( 3 x − 4 y − 2 z ) 2 − ( 3 x + 4 y − 2 z ) 2 = 9 x 2 + 16 y 2 + 4 z 2 − 24 x y + 16 y z − 12 x z − 9 x 2 − 16 y 2 − 4 z 2 − 24 x y + 16 y z + 12 x z = 32 y z − 48 x y \begin{aligned} \\ \therefore\left(3x-4y-2z\right)^2-\left(3x+4y-2z\right)^2=\;&\cancel{9x^2}+\cancel{16y^2}+\cancel{4z^2}-24xy+16yz-\cancel{12xz}- \\ &\cancel{9x^2}-\cancel{16y^2}-\cancel{4z^2}-24xy+16yz+\cancel{12xz} \\ =&\;32yz-48xy\end{aligned} ∴ ( 3 x − 4 y − 2 z ) 2 − ( 3 x + 4 y − 2 z ) 2 = = 9 x 2 + 1 6 y 2 + 4 z 2 − 2 4 x y + 1 6 y z − 1 2 x z − 9 x 2 − 1 6 y 2 − 4 z 2 − 2 4 x y + 1 6 y z + 1 2 x z 3 2 y z − 4 8 x y
उदाहरण 10: ( 7 p − 6 q + 5 r ) 2 + ( 7 p + 6 q − 5 r ) 2 \left(7p-6q+5r\right)^2+\left(7p+6q-5r\right)^2 ( 7 p − 6 q + 5 r ) 2 + ( 7 p + 6 q − 5 r ) 2 उत्तर: ∴ ( 7 p − 6 q + 5 r ) 2 = ( 7 p ) 2 + ( − 6 q ) 2 + ( 5 r ) 2 + 2 ( ( 7 p ) × ( − 6 q ) ) + 2 ( ( − 6 q ) × ( 5 r ) ) + 2 ( ( 7 p ) × ( 5 r ) ) = 49 p 2 + 36 q 2 + 25 r 2 − 84 p q − 60 q r + 70 p r \begin{aligned} \\ \therefore\left(7p-6q+5r\right)^2=&\left(7p\right)^2+\left(-6q\right)^2+\left(5r\right)^2+2\left(\left(7p\right)\times \left(-6q\right)\right)+ \\ &\;2\left(\left(-6q\right)\times \left(5r\right)\right)+2\left(\left(7p\right)\times \left(5r\right)\right) \\ =&\;49p^2+36q^2+25r^2-84pq-60qr+70pr\end{aligned} ∴ ( 7 p − 6 q + 5 r ) 2 = = ( 7 p ) 2 + ( − 6 q ) 2 + ( 5 r ) 2 + 2 ( ( 7 p ) × ( − 6 q ) ) + 2 ( ( − 6 q ) × ( 5 r ) ) + 2 ( ( 7 p ) × ( 5 r ) ) 4 9 p 2 + 3 6 q 2 + 2 5 r 2 − 8 4 p q − 6 0 q r + 7 0 p r ∴ ( 7 p + 6 q − 5 r ) 2 = ( 7 p ) 2 + ( 6 q ) 2 + ( − 5 r ) 2 + 2 ( ( 7 p ) × ( 6 q ) ) + 2 ( ( 6 q ) × ( − 5 r ) ) + 2 ( ( 7 p ) × ( − 5 r ) ) = 49 p 2 + 36 q 2 + 25 r 2 + 84 p q − 60 q r − 70 p r \begin{aligned} \\ \therefore\left(7p+6q-5r\right)^2=&\left(7p\right)^2+\left(6q\right)^2+\left(-5r\right)^2+2\left(\left(7p\right)\times \left(6q\right)\right)+ \\ &\;2\left(\left(6q\right)\times \left(-5r\right)\right)+2\left(\left(7p\right)\times \left(-5r\right)\right) \\ =&\;49p^2+36q^2+25r^2+84pq-60qr-70pr\end{aligned} ∴ ( 7 p + 6 q − 5 r ) 2 = = ( 7 p ) 2 + ( 6 q ) 2 + ( − 5 r ) 2 + 2 ( ( 7 p ) × ( 6 q ) ) + 2 ( ( 6 q ) × ( − 5 r ) ) + 2 ( ( 7 p ) × ( − 5 r ) ) 4 9 p 2 + 3 6 q 2 + 2 5 r 2 + 8 4 p q − 6 0 q r − 7 0 p r ∴ ( 7 p − 6 q + 5 r ) 2 + ( 7 p + 6 q − 5 r ) 2 = 49 p 2 + 36 q 2 + 25 r 2 − 84 p q − 60 q r + 70 p r + 49 p 2 + 36 q 2 + 25 r 2 + 84 p q − 60 q r − 70 p r = 98 p 2 + 72 q 2 + 50 r 2 − 120 q r \begin{aligned} \\ \therefore\left(7p-6q+5r\right)^2+\left(7p+6q-5r\right)^2=&\;49p^2+36q^2+25r^2-\cancel{84pq}-60qr+\cancel{70pr}+ \\ &\;49p^2+36q^2+25r^2+\cancel{84pq}-60qr-\cancel{70pr} \\ =&\;98p^2+72q^2+50r^2-120qr\end{aligned} ∴ ( 7 p − 6 q + 5 r ) 2 + ( 7 p + 6 q − 5 r ) 2 = = 4 9 p 2 + 3 6 q 2 + 2 5 r 2 − 8 4 p q − 6 0 q r + 7 0 p r + 4 9 p 2 + 3 6 q 2 + 2 5 r 2 + 8 4 p q − 6 0 q r − 7 0 p r 9 8 p 2 + 7 2 q 2 + 5 0 r 2 − 1 2 0 q r
थोडक्यात: ( x + y ) 2 = x 2 + 2 x y + y 2 (x+y)^2 = x^{2} + 2xy + y^{2} ( x + y ) 2 = x 2 + 2 x y + y 2 ( x − y ) 2 = x 2 − 2 x y + y 2 (x-y)^2 = x^2 - 2xy + y^2 ( x − y ) 2 = x 2 − 2 x y + y 2 ( x + y ) ( x − y ) = x 2 − x y + x y − y 2 = x 2 − y 2 (x+y)(x-y) = x^2 - xy + xy - y^2 = x^2 - y^2 ( x + y ) ( x − y ) = x 2 − x y + x y − y 2 = x 2 − y 2 ( a + b ) 3 \mathbf{\left(a+b\right)^3} ( a + b ) 3 ( a − b ) 3 \mathbf{\left(a-b\right)^3} ( a − b ) 3 ( a + b + c ) 2 \mathbf{\left(a+b+c\right)^2} ( a + b + c ) 2
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इयत्ता 8 वीचे पाठयपुस्तक: इथे क्लिक करा
