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

$\text{Here,}(a^2-b^2)+(b^2-c^2)+(c^2-a^2)=0$

$\therefore (a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-a^2{)}^3=3(a^2-b^2\left)\right(b^2-c^2\left)\right(c^2-a^2)$

$\text{Also,}(a-b)+(b-c)+(c-a)=0$

$\therefore (a-b{)}^3+(b-c{)}^3+(c-a{)}^3=3(a-b\left)\right(b-c\left)\right(c-a)$

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

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

$=(a+b)(b+c)(c+a).$