Question

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

Answer 1

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

We know that,
If $x+y+z=0$ then $x^3+y^3+z^3=3xyz$

Consider, $x=a^2-b^2,y=b^2-c^2$ and $z=c^2-a^2$
Since, $a^2-b^2+b^2-c^2+c^2-a^2=0,$

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

Also, $a-b+b-c+c-a=0$
$\therefore (a-b{)}^3+(b-c{)}^3+(c-a{)}^3$
$=3(a-b)(b-c)(c-a)$

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

Hence, $\left(d\right)$ is the correct option.