Answer
$\frac{x^{2}+2x+4}{3x}; x\ne-2,0,2;$
Work Step by Step
$\frac{x^{3}-8}{x^{2}-4}. \frac{(x+2)}{3x}$
In the given expression $(x^{3}-8)$ can be written as $(x^{3}-2^{3})$
It is in the form of $(A^{3}-B^{3})$.
$(x^{3}-8)$ can be factored using the formula $(A^{3}-B^{3}) = (A-B)(A^{2}+AB+B^{2})$. So,
$(x^{3}-2^{3}) = (x-2)(x^{2}+2x+4)$.
$(x^{2}-4) = x^{2}-2^{2}$ is the difference of squares and the factors are $(x+2)(x-2)$
$\frac{x^{3}-8}{x^{2}-4}. \frac{x+2}{3x} = \frac{(x-2)(x^{2}+2x+4)}{(x+2)(x-2)} . \frac{(x+2)}{3x}$
The values 2, -2, 0 for $x$ makes the denominator zero. So, we must exclude them.
$ = \frac{(x-2)(x^{2}+2x+4)}{(x+2)(x-2)} . \frac{(x+2)}{3x} ; x\ne -2,0,2$
Divide out common factors and the result is
$ = \frac{(x^{2}+2x+4)}{3x} ; x\ne -2,0,2$
