Answer
$\{2,-1-\sqrt{3}i,-1+\sqrt{3}i\}$
Work Step by Step
Recognize a difference of cubes,
$x^{3}-2^{3}=(x-2)(x^{2}+2x+2^{2})$
One solution is $x=2$.
For the other two, solve
$x^{2}+2x+4=0$
Use the quadratic formula
$a=1, b=2, c=4$
$x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$x=\displaystyle \frac{-2\pm\sqrt{4-16}}{2}=\frac{-2\pm\sqrt{-12}}{2}$
...If $N$ is a positive real number, we define the principal square root of $-N,$ denoted by $\sqrt{-N},$ as $\sqrt{-N}=\sqrt{N}i$
$=\displaystyle \frac{-2\pm i\sqrt{12}}{2}$
$=-\displaystyle \frac{2}{2}\pm\frac{2\sqrt{3}}{2}i$
$=-1\pm\sqrt{3}i$
Solution set = $\{2,-1-\sqrt{3}i,-1+\sqrt{3}i\}$
