Answer
$ \frac{1}{6} $.
Work Step by Step
The given expression is
$\frac{1}{x^2-2x-8}\div\left( \frac{1}{x-4}-\frac{1}{x+2} \right)$
The LCM of the denominators in the bracket is =(x+2)(x-4).
Multiply the fractions in the bracket to form the LCD in the denominator.
$\frac{1}{x^2-2x-8}\div\left( \frac{(x+2)}{(x+2)(x-4)}-\frac{(x-4)}{(x+2)(x-4)} \right)$
Add numerators in the bracket.
$\frac{1}{x^2-2x-8}\div\left( \frac{(x+2)-(x-4)}{(x+2)(x-4)} \right)$
Simplify the bracket.
$\frac{1}{x^2-2x-8}\div\left( \frac{x+2-x+4}{(x+2)(x-4)} \right)$
$\frac{1}{x^2-2x-8}\div\left( \frac{6}{(x+2)(x-4)} \right)$
Invert the divisor and multiply.
$\frac{1}{x^2-2x-8}\cdot\frac{(x+2)(x-4)}{6} $
Factor the denominator of the first fraction.
$\Rightarrow x^2-2x-8$
Rewrite the middle term $-2x$ as $-4x+2x$
$\Rightarrow x^2-4x+2x-8$
Group the terms.
$\Rightarrow (x^2-4x)+(2x-8)$
Factor each group.
$\Rightarrow x(x-4)+2(x-4)$
Factor out $(x-4)$.
$\Rightarrow (x-4)(x+2)$
Substitute the factor into the fraction.
$\Rightarrow \frac{1}{(x-4)(x+2)}\cdot\frac{(x+2)(x-4)}{6} $
Cancel common terms.
$\Rightarrow \frac{1}{6} $.
