Answer
$\displaystyle \{-3,\frac{3}{2}-\sqrt{6}i,\frac{3}{2}+\sqrt{6}i\}$
Work Step by Step
Recognize a sum of cubes,
$x^{3}+3^{3}=(x+3)(x^{2}-3x+3^{2})$
One solution is $x=-3$
For the other two, solve
$x^{2}-3x+9=0$
Use the quadratic formula
$a=1, b=-3, c=9$
$x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$x=\displaystyle \frac{3\pm\sqrt{9-36}}{2}=\frac{3\pm\sqrt{-24}}{2}$
...If $N$ is a positive real number, we define the principal square root of $-N,$ denoted by $\sqrt{-N},$ as $\sqrt{-N}=\sqrt{N}i$
$=\displaystyle \frac{3\pm i\sqrt{24}}{2}$
$=\displaystyle \frac{3}{2}\pm\frac{2\sqrt{6}}{2}i$
$=\displaystyle \frac{3}{2}\pm\sqrt{6}i$
Solution set = $\displaystyle \{-3,\frac{3}{2}-\sqrt{6}i,\frac{3}{2}+\sqrt{6}i\}$
