Question

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Answer 1

Step-by-step explanation:

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

Let in numerator

x=(a2−b2),y=(b2−c2),z=(c2−a2)

⟹x+y+z=0(1)

We know, (x+y+z)3=x3+y3+z3+3(x+y+z)(xy+yz+zx)−3xyz

0=x3+y3+z3+3(0)(xy+yz+zx)−3xyz

x3+y3+z3=3xyz(2)

Let in denominator,

p=(a−b),q=(b−c),r=(c−a)

⟹p+q+r=0

⟹

p3+q3+r3=3pqr(3)

(a2−b2)3+(b2−c2)3+(c2−a2)3(a−b)3+(b−c)3+(c−a)3=3xyz3pqr

=(a2−b2)(b2−c2)(c2−a2)(a−b)(b−c)(c−a)

=(a−b)(b−c)(c−a)(a+b)(b+c)(c+a)(a−b)(b−c)(c−a)

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

Answer 2

Answer:

here is the solution go through this and I hope it helps ...