Question

Simplify : $(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-a^2{)}^3÷[(a-b{)}^3+(b-c{)}^3+(c-a{)}^3]$

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 $x^3+y^3+z^3=3xyz$

Since,

$\Rightarrow a^2-b^2+b^2-c^2+c^2-a^2=0$, therefore,

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

Also,

$\Rightarrow a-b+b-c+c-a=0$

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

Therefore,

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

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

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

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