Answer
$-\frac{81}{2}+\frac{81\sqrt {3}}{2}i$
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
$[3(\cos \frac{\pi}{6}+i\sin \frac{\pi}{6})]^{4}=(3)^{4}(\cos 4\cdot\frac{\pi}{6}+i\sin 4\cdot\frac{\pi}{6})$
$=81(\cos \frac{2\pi}{3}+i\sin\frac{2\pi}{3})$
In standard form, our result is
$=81(-\frac{1}{2}+i\cdot\frac{\sqrt 3}{2})$
$=-\frac{81}{2}+\frac{81\sqrt {3}}{2}i$