Answer
$32+32i\sqrt {3}$
Work Step by Step
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$[2(\cos 10^{\circ}+i\sin 10^{\circ})]^{6}=(2)^{6}(\cos 6\cdot10^{\circ}+i\sin 6\cdot10^{\circ})$
$=64(\cos 60^{\circ}+i\sin60^{\circ})$
In standard form, our result is
$=64(\frac{1}{2}+i\cdot\frac{\sqrt 3}{2})$
$=32+32i\sqrt {3}$
