simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $? How can I simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $ to rectangular form $z = a+bi$?
(Note: Wolfram Alpha says the answer is $z=-8$. My professor says the answer is $z=\pm8$.)
I've tried to figure this out for a couple hours now, but I'm getting nowhere. 
Any help is much appreciated!

$\:\sqrt{-2+2\sqrt{-3}}\:$ can be denested by a radical denesting formula that I discovered as a teenager.

Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \  \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $  
and, furthermore,  $\rm\:w\:$ has trace $\rm\: =\: w+w' =  (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\:  2\:a$

Here $ {-}2+2\sqrt{-3}\:$ has norm $= 16.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = -4\ $ yields $\  2+2\sqrt{-3}\:$
and this has $\rm\ \sqrt{trace}\: =\: 2,\ \ hence\ \  \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:1+\sqrt{-3}.$
Checking we have
$\ \smash[t]{\displaystyle \left(1+\sqrt{-3}\right)^2 =\ 1-3 + 2\sqrt{-3}\ =\ -2 + 2 \sqrt{-3}}$
Therefore $\quad\ \begin{eqnarray}\rm\:(-2 + 2\sqrt{-3})^{3/2} &=&\ (-2+2\sqrt{-3})\ (-2+2\sqrt{-3})^{1/2} \\  &=&\ -2\,(1-\sqrt{-3})\ (1+\sqrt{-3}) \\
 &=&\ -8\rm\ \ \ (up\ to\ sign) 
\end{eqnarray}$
See this answer for general radical denesting algorithms.
