#### Answer

$\displaystyle \frac{x+5}{x^{3}-1}\quad $or$ \displaystyle \quad\frac{x+5}{(x-1)(x^{2}+x+1)}$

#### Work Step by Step

$x^{3}-1$ is a difference of cubes,
$x^{3}-1^{3}=(x-1)(x^{2}+x+1)$
$x^{2}+x+1$ has no real zeros as $b^{2}-4ac=1-4$ ... is negative.
LCD = $(x-1)(x^{2}+x+1)=x^{3}-1$
Write all fractions with the LCD,
$\displaystyle \frac{2}{x^{3}-1}+\frac{4}{x^{3}-1}+\frac{1}{x^{2}+x+1}\cdot\frac{x-1}{x-1}=$
$=\displaystyle \frac{2+4+x-1}{x^{3}-1}$
$=\displaystyle \frac{x+5}{x^{3}-1}\quad $or$ \displaystyle \quad\frac{x+5}{(x-1)(x^{2}+x+1)}$