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Question

Find (a2−b2)3+(b2−c2)3+(c2−a2)3(a−b)3+(b−c)3+(c−a)3=

A

3(a+b)(b+c)(c+a)

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B

3(a-b)(b-c)(c-a)

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C

(a-b)(b-c)(c-a)

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D

(a+b)(b+c)(c+a)

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Solution

Since, we have if x+y+z=0 then x3+y3+z3=3xyz Put x=a−b,y=b−c,z=c−a then x+y+z=a−b+b−c+c−a=0⇒(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a).....(i)Similar way, if x=a2−b2,y=b2−c2,z=c2−a2 then x+y+z=a2−b2+b2−c2+c2−a2=0Thus, (a2−b2)3+(b2−c2)3+(c2−a2)3=3(a2−b2)(b2−c2)(c2−a2)....(ii)Now consider given expression =(a2−b2)3+(b2−c2)3+(c2−a2)3(a−b)3+(b−c)3+(c−a)3Substitute equation (i) and equation (ii) in above expression, we get =3(a2−b2)(b2−c2)(c2−a2)3(a−b)(b−c)(c−a) [Since, (a2−b2)=(a+b)(a−b)]=3(a+b)(a−b)(b+c)(b−c)(c+a)(c−a)3(a−b)(b−c)(c−a)=(a+b)(b+c)(c+a)Therefore, the value of the given expression is (a+b)(b+c)(c+a)

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