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Class 10 > Factorise: (a -- b)3 + (b -- c)3 + (c -- a)3
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Factorise: (a – b)3 + (b – c)3 + (c – a)3

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3(a-b)(b-c)(c-a)
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3(a-b+b-c+c-a)= 3*(0)=0
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First factorize (a-b)^3 , (b-c)^3 and (c-a)^3 separately and then add them together to get the final result. (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 ................(1) (b-c)^3 = b^3 - c^3 - 3b^2c + 3bc^2 .................(2) (c-a)^3 = c^3 - a^3 - 3ac^2 + 3 a^2c ...
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First factorize (a-b)^3 , (b-c)^3 and (c-a)^3 separately and then add them together to get the final result. (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 ................(1) (b-c)^3 = b^3 - c^3 - 3b^2c + 3bc^2 .................(2) (c-a)^3 = c^3 - a^3 - 3ac^2 + 3 a^2c ..................(3) Now add (1), (2) and (3). This gives, a^3 - b^3 - 3a^2b + 3ab^2 + b^3 - c^3 - 3b^2c + 3bc^2 + c^3 - a^3 - 3ac^2 + 3 a^2c = -3a^2b + 3ab^2 - 3b^2c + 3bc^2 - 3ac^2 + 3 a^2c read less
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a3-3a2b+3ab2-b3 + b3-3b2c+3bc2-c3 + c3-3c2a+3ca2-a3 = -3a2b+3ab2 -3b2c+3bc2 -3c2a+3ca2 = 3a2 ( c-b) + 3b2 ( a-c ) + 3c2 (b-a)
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(a-b)(b-c)(c-a)
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we know, x3+y3=(x+y)(x2−xy+y2)x3+y3=(x+y)(x2−xy+y2) so, (a−b)3+(b−c)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2) (a−b)3+(b−c)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2) now, (a−b)3+(b−c)3+(c−a)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2)+(c−a)3=(c−a)(−(a−b)2+(a−b)(b−c)−(b−c)2+(c−a)2) (a−b)3+(b−c)3+(c−a)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2)+(c−a)3=(c−a)(−(a−b)2+(a−b)(b−c)−(b−c)2+(c−a)2) now,...
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we know, x3+y3=(x+y)(x2−xy+y2)x3+y3=(x+y)(x2−xy+y2) so, (a−b)3+(b−c)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2) (a−b)3+(b−c)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2) now, (a−b)3+(b−c)3+(c−a)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2)+(c−a)3=(c−a)(−(a−b)2+(a−b)(b−c)−(b−c)2+(c−a)2) (a−b)3+(b−c)3+(c−a)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2)+(c−a)3=(c−a)(−(a−b)2+(a−b)(b−c)−(b−c)2+(c−a)2) now, (c−a)2−(a−b)2=(c−a+a−b)(c−a−a+b)=(c−b)(c−2a+b)(c−a)2−(a−b)2=(c−a+a−b)(c−a−a+b)=(c−b)(c−2a+b) the expression becomes, (c−a)((c−b)(c−2a+b)+(b−c)(a−2b+c))=(c−a)(b−c)(−c+2a−b+a−2b+c) =3(c−a)(b−c)(a−b) read less
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ans is zero
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We know a^3 + b^3 +c^3= 3abc if a+b+c=0...so answer is 3(a-b)(b-c)(c-a)
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3(a-b)(b-c)(c-a)
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