Answer
$-\frac{6}{5}$.
Work Step by Step
The given expression is
$=\frac{\frac{3}{x+1}-\frac{3}{x-1}}{\frac{5}{x^2-1}}$
Factor the term $x^2-1$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=x^2-1^2$
$=(x+1)(x-1)$
$=\frac{\frac{3}{x+1}-\frac{3}{x-1}}{\frac{5}{(x+1)(x-1)}}$
Multiply the numerator and the denominator by $(x+1)(x-1)$.
$=\frac{(x+1)(x-1)}{(x+1)(x-1)}\cdot \frac{\frac{3}{x+1}-\frac{3}{x-1}}{\frac{5}{(x+1)(x-1)}}$
Use the distributive property.
$=\frac{(x+1)(x-1) \cdot \frac{3}{x+1}-(x+1)(x-1) \cdot \frac{3}{x-1}}{(x+1)(x-1) \cdot\frac{5}{(x+1)(x-1)}}$
Simplify.
$=\frac{3(x-1) -3(x+1)}{5}$
$=\frac{3x-3 -3x-3}{5}$
$=\frac{-6}{5}$
$=-\frac{6}{5}$.
