Question

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

Answer 1

let $a^2-b^2=x_1;b^2-c^2=y_1;c^2-a^2=z_1$

$x_1+y_1+z_1=0$

$\Rightarrow x_1^3+y_1^3+z_1^3=3x_1{y}_1{z}_1=3(a^2-b^2)(b^2-c^2)(c^2-a^2)$

let $a-b=x_2;b-c=y_2;c-a=z_2$

$x_2+y_2+z_2=0$

$\Rightarrow x_2^3+y_2^3+z_2^3=3x_2{y}_2{z}_2=3(a-b)(b-c)(c-a)$

So,

$\frac{x_1^3+y_1^3+z_1^3}{x_2^3+y_2^3+z_2^3}=(a+b)(b+c)(c+a)$