Question

please answer the question.​

Answer 1

We know that x^3+y^3+z^3=3xyz

if x+y+z=0.

Now (a^2-b^2 ) + ( b^2 -c^2)+(c^2-a^2)

a^2-a^2+b^2-b^2+c^2-c^2

=0.

(a^2-b^2)^3+(b^2-c^2)^3 +(c^2-a^2)^3=

3(a^2-b^2)(b^2-c^2)(c^2-a^2)

Similarly we can say that

(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a).

therefore, the question becomes

3(a^2-b^2)(b^2-c^2)(c^2-a^2)/3(a-b)(b-c)(c-a)

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){common terms get cancelled}

this is the required answer.

Hope it helped.