Question

Simplify:
$\frac{m^2-n^2}{{\left(m+n\right)}^2}\times \frac{m^2+mn+n^2}{m^3-n^3}$

Answer 1

Let $I=\frac{m^2-n^2}{{(m+n)}^2}\times \frac{m^2+mn+n^2}{m^3-n^3}$

Using identity-

$a^3-b^3=(a-b)(a^2+b^2+ab)$

$a^2-b^2=(a-b)(a+b)$

$\therefore I=\frac{(m-n)(m+n)}{{(m+n)}^2}\times \frac{(m^2+mn+n^2)}{(m-n)(m^2+mn+n^2)}$

$\Rightarrow I=\frac{1}{(m+n)}$

Hence $\frac{m^2-n^2}{{(m+n)}^2}\times \frac{m^2+mn+n^2}{m^3-n^3}=\frac{1}{(m+n)}$.