Answer
$\displaystyle \frac{-2(x-4)}{x^{3}-8}$ or $\displaystyle \frac{-2(x-4)}{(x-2)(x^{2}+2x+4)}$
Work Step by Step
$x^{3}-8$ is a difference of cubes,
$x^{3}-2^{3}=(x-2)(x^{2}+2x+4)$
$x^{2}+2x+4$ has no real zeros as $b^{2}-4ac=4-8$ ... is negative.
LCD = $(x-2)(x^{2}+2x+4)$
Exclude solutions which yield 0 in denominators:
$x\not\in\{2\}\qquad (*)$
Write all fractions with the LCD,
$\displaystyle \frac{1}{x^{3}-8}+\frac{3}{x^{3}-8}-\frac{2}{x^{2}+2x+4}\cdot\frac{x-2}{x-2}$
$=\displaystyle \frac{1+3-2(x-2)}{x^{3}-8}$
$=\displaystyle \frac{4-2x+4}{x^{3}-8}$
$=\displaystyle \frac{8-2x}{x^{3}-8}$
$=\displaystyle \frac{-2(x-4)}{x^{3}-8}$ or $\displaystyle \frac{-2(x-4)}{(x-2)(x^{2}+2x+4)}$
