Answer
$-32\sqrt{3}+32i$
Work Step by Step
DeMoivre’s Theorem:
For any complex number $z=r\left( \cos \theta +i\sin \theta \right)$, if n is a positive integer $\left( z>0 \right)$ then,
$\begin{align}
& {{z}^{n}}={{\left[ r\left( \cos \theta +i\sin \theta \right) \right]}^{n}} \\
& ={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)
\end{align}$
So, the given complex number can be written as:
$\begin{align}
& {{\left[ 4\left( \cos 50{}^\circ +i\sin 50{}^\circ \right) \right]}^{3}}={{4}^{3}}\left[ \cos 3\left( 50{}^\circ \right)+i\sin 3\left( 50{}^\circ \right) \right] \\
& =64\left( \cos 150{}^\circ +i\sin 150{}^\circ \right) \\
& =64\left( -\frac{\sqrt{3}}{2}+i\frac{1}{2} \right) \\
& =-32\sqrt{3}+32i
\end{align}$
Hence, ${{\left[ 4\left( \cos 50{}^\circ +i\sin 50{}^\circ \right) \right]}^{3}}=-32\sqrt{3}+32i$.
