The value of $\frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}$ is equal to :

$\frac{1}{(a+b)(b+c)(c+a)}$

$\frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}$ = ?

When x³ + y³ + z³ = 0 then, x³ + y³ + z³ = 3xyz   

and we also know that,

a² - b² = (a + b) (a – b)

Applying this condition in the given question,

$\frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}$ = $\frac{3(a-b)(b-c)(c-a)}{3(a^2-b^2)(b^2-c^2)(c^2-a^2)}$

= $\frac{3(a-b)(b-c)(c-a)}{3(a-b)(b-c)(c-a)(a-b)(b-c)(c-a)}$

= $\frac{1}{(a+b)(b+c)(c+a)}$