Answer
$16+16i$
Work Step by Step
$-2+2i$ in trigonometric form is
$-2+2i=2\sqrt {2}(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4})$
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$(-2+2i)^{3}=[2\sqrt {2}(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4})]^{3}$
$=(2\sqrt {2})^{3}(\cos 3\cdot\frac{3\pi}{4}+i\sin 3\cdot\frac{3\pi}{4})$
$=16\sqrt {2}(\cos \frac{9\pi}{4}+i\sin\frac{9\pi}{4})$
$=16\sqrt {2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$
($\frac{\pi}{4}$ and $\frac{9\pi}{4}$ are coterminal.)
In standard form, our result is
$=16\sqrt {2}(\frac{\sqrt 2}{2}+i\cdot\frac{\sqrt 2}{2})$
$=16+16i$