Answer
$\frac{1}{6}; x \ne 4,-2;$
Work Step by Step
$\frac{1}{x^{2}-2x-8} \div ( \frac{1}{x-4} - \frac{1}{x+2} )$
Factors of $x^{2}-2x-8$ are $(x-4),(x+2)$
$= \frac{1}{(x-4)(x+2)} \div ( \frac{1}{x-4} - \frac{1}{x+2} ); x \ne 4,-2;$
Taking LCD in the Divisor part,
$= \frac{1}{(x-4)(x+2)} \div ( \frac{(x+2)-(x-4)}{(x-4)(x+2)} ); x \ne 4,-2;$
$= \frac{1}{(x-4)(x+2)} \div ( \frac{x+2-x+4}{(x-4)(x+2)} ); x \ne 4,-2;$
$= \frac{1}{(x-4)(x+2)} \div ( \frac{6}{(x-4)(x+2)} ); x \ne 4,-2;$
$= \frac{1}{(x-4)(x+2)} \times ( \frac{(x-4)(x+2)}{6} ); x \ne 4,-2;$
Cross out common factors.
$= \frac{1}{6}; x \ne 4,-2;$
