Answer
The three cube roots are:
\[
\begin{array}{c}
[\cos 10+i \sin 10]2 \\
[\cos 130+i \sin 130] 2\\
[\cos 250+i \sin 250]2
\end{array}
\]
Work Step by Step
\[
4 \sqrt{3}+4 i
\]
The modulus of a complex number in the standard form $x+y i$ is:
\[
\begin{array}{c}
\sqrt{y^{2}+x^{2}}=r \\
\sqrt{4^{2}+(4 \sqrt{3})^{2}}=r \\
\sqrt{16+48}=r \\
\sqrt{64} =r\\
8=r
\end{array}
\]
The $\theta$ angle is the least positive angle for which:
\[
\frac{\sqrt{3}}{2} =\frac{4 \sqrt{3}}{8}=\frac{x}{r}=\cos \theta
\]
,
\[
\frac{1}{2}=\frac{4}{8}=\frac{y}{r}=\sin \theta
\]
Thus
\[
30^{\circ}=\theta
\]
Using the definition of a complex number in trigonometric form:
\[
\begin{aligned}
&=(\cos \theta+i \sin \theta)r \\
&=\left(\cos 30^{\circ}+i \sin 30^{\circ}\right)8
\end{aligned}
\]
Using the roots theorem:
\[
\left[\cos \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)+i \sin \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)\right]r \frac{1}{n}=w_{k}
\]
When $0=k$
\[
\left[\cos \left(\frac{30^{\circ}}{3}+\frac{360^{\circ}}{3}(0)\right)+i \sin \left(\frac{30^{\circ}}{3}+\frac{360^{\circ}}{3}(0)\right)\right]8 \frac{1}{3}=w_{0}
\]
Simplify:
\[
={2\left[\cos 10^{\circ}+i \sin 10^{\circ}\right]}
\]
When $1=k$ :
\[
\left[\cos \left(\frac{30^{\circ}}{3}+\frac{360^{\circ}}{3}(1)\right)+i \sin \left(\frac{30^{\circ}}{3}+\frac{360^{\circ}}{3}(1)\right)\right]8^{\frac{1}{3}}=w_{1}
\]
Simplify:
\[
\begin{array}{r}
=\left[\cos \left(10^{\circ}+120^{\circ}\right)+i \sin \left(10^{\circ}+120^{\circ}\right)\right] 2\\
=\left[2\left[\cos 130^{\circ}+i \sin 130^{\circ}\right]\right.
\end{array}
\]
When $2=k$ :
\[
\left[\cos \left(\frac{30^{\circ}}{3}+\frac{360^{\circ}}{3}(2)\right)+i \sin \left(\frac{30^{\circ}}{3}+\frac{360^{\circ}}{3}(2)\right)\right]8^{\frac{1}{3}}=w_{2}
\]
Simplify:
\[
\begin{aligned}
=\left[\cos \left(10^{\circ}+240^{\circ}\right)+i \sin \left(10^{\circ}+240^{\circ}\right)\right]2 & \\
=&\left[2\left[\cos 250^{\circ}+i \sin 250^{\circ}\right]\right.
\end{aligned}
\]
