Answer
The three cube roots are:
\[
\begin{array}{c}
[\cos 50+i \sin 50]2 \\
[\cos 170+i \sin 170]2 \\
[\cos 290+i \sin 290]2
\end{array}
\]
Work Step by Step
\[
-4 \sqrt{3}+4 i
\]
The modulus of a complex mumber in the standard form $x+y i$ is:
\[
\begin{array}{c}
\sqrt{y^{2}+x^{2}}=r \\
\sqrt{(-4 \sqrt{3})^{2}+4^{2}} =r\\
\sqrt{16+48}=r \\
\sqrt{64}=r \\
8=r
\end{array}
\]
The $\theta$ angle is the least positive angle for which:
\[
-\frac{\sqrt{3}}{2}=\frac{-4 \sqrt{3}}{8}=\frac{x}{r}=\cos \theta
\]
,
\[
\frac{1}{2}=\frac{4}{8}=\frac{y}{r}=\sin \theta
\]
$\mathrm{Thus}$
\[
150^{\circ}=\theta
\]
Using the definition of the trigonometric form of a complex number:
\[
\begin{aligned}
&=(\cos \theta+i \sin \theta)r \\
&=\left(\cos 150^{\circ}+i \sin 150^{\circ}\right)8
\end{aligned}
\]
Using the roots theorem:
\[
w_{k}=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)+i \sin \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)\right]
\]
When $0=k$ :
\[
\left[\cos \left(\frac{150^{\circ}}{3}+\frac{360^{\circ}}{3}(0)\right)+i \sin \left(\frac{150^{\circ}}{3}+\frac{360^{\circ}}{3}(0)\right)\right]8^{\frac{1}{3}}=u_{0}
\]
Simplify:
\[
=2
\]
$-i \sin 50^{\circ}$
When $1=k$:
\[
\left[\cos \left(\frac{150^{\circ}}{3}+\frac{360^{\circ}}{3}(1)\right)+i \sin \left(\frac{150^{\circ}}{3}+\frac{360^{\circ}}{3}(1)\right)\right]8^{\frac{1}{3}}=w_{1}
\]
Simplify:
\[
\begin{array}{c}
=\left[\cos \left(50^{\circ}+120^{\circ}\right)+i \sin \left(50^{\circ}+120^{\circ}\right)\right] 2\\
=\left[2\left[\cos 170^{\circ}+i \sin 170^{\circ}\right]\right.
\end{array}
\]
When $2=k$:
\[
\left[\cos \left(\frac{150^{\circ}}{3}+\frac{360^{\circ}}{3}(2)\right)+i \sin \left(\frac{150^{\circ}}{3}+\frac{360^{\circ}}{3}(2)\right)\right]8^{\frac{1}{3}}=w_{2}
\]
Simplify:
\[
\begin{aligned}
=\left[\cos \left(50^{\circ}+240^{\circ}\right)+i \sin \left(50^{\circ}+240^{\circ}\right)\right]2 & \\
=& 2\left[\cos 290^{\circ}+i \sin 290^{\circ}\right]
\end{aligned}
\]
