Answer
Solution set: $\{-5\}$
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 = $(x-1)(x^{2}+x+1)$
Exclude solutions which yield 0 in denominators:
$x\not\in\{1\}\qquad (*)$
Multiply with the LCD, $(x-1)(x^{2}+x+1)=x^{3}-1$
$2+4=-1(x-1)$
$6=-x+1$
$5=-x$
$ x=-5\qquad$ ...satisfies (*).
Solution set: $\{-5\}$
