Question

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

Answer 1

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

$=\frac{1}{a-b}(\frac{a+3b}{a-b}+\frac{a-3b}{a+b})$

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

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

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