Answer
$=\displaystyle \frac{x^{2}+2x+4}{3x},\qquad x\neq-2,0,2$
Work Step by Step
Factor what you can:
$ x^{3}-8= \quad$...difference of cubes = $x^{3}-2^{3},$
$=(x-2)(x^{2}+2x+4)$
$x^{2}-4 =$ difference of squares $= (x-2)(x+2)$
$x+2$ is factored
$3x$ is factored
Expression $ =\displaystyle \frac{(x-2)(x^{2}+2x+4)}{(x-2)(x+2)}\cdot\frac{(x+2)}{3x} $
$\qquad$
...exclude the values that yield 0 in the denominator:
$ x\neq -2,0,2$
... cancel common factors:
$=\displaystyle \frac{x^{2}+2x+4}{3x},\qquad x\neq-2,0,2$
