Answer
Solution set: $\{4\}$
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:
$x\not\in\{2\}\qquad (*)$
Multiply with the LCD, $(x-2)(x^{2}+2x+4)=x^{3}-8$
$1+3=2(x-2)$
$4=2x-4$
$8=2x$
$ x=4\qquad$ ...satisfies (*).
Solution set: $\{4\}$
