Answer
The three cube roots are:
\[
\begin{array}{r}
\left[\cos 30^{\circ}+i \sin 30^{\circ}\right] 4\\
4 [\cos 150^{\circ}+i \sin 150^{\circ}] \\
\left[\cos 270^{\circ}+i \sin 270^{\circ}\right]4
\end{array}
\]
Work Step by Step
\[
64 i+0=64 i
\]
The modulus of a complex number in the standard form $x+y i$ is:
\[
\begin{array}{l}
\sqrt{y^{2}+x^{2}}=r \\
\sqrt{0^{2}+64^{2}} =r\\
\sqrt{4096}=r \\
64=r
\end{array}
\]
The $\theta$ angle is the least positive angle for which:
\[
0=\frac{0}{64}=\frac{x}{r}=\cos \theta
\]
,
\[
1=\frac{64}{64}=\frac{y}{r}=\sin \theta
\]
Thus,
\[
90^{\circ}=\theta
\]
Using the definition of the trigonometric form of a complex number:
\[
\begin{aligned}
&=(\cos \theta+i \sin \theta)r \\
=&\left(\cos 90^{\circ}+i \sin 90^{\circ}\right)64
\end{aligned}
\]
Using the roots theorem:
\[
\left[\cos \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)+i \sin \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)\right]r^{\frac{1}{n}}=u_{k}
\]
$0=\mathrm{W} \log k$
\[
\left[\cos \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(0)\right)+i \sin \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(0)\right)\right]64^{\frac{1}{3}}=u_{0}
\]
Simplify:
\[
=\sqrt{4\left[\cos 30^{\circ}+i \sin 30^{\circ}\right]}
\]
When $1=k$ :
\[
\left[\cos \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(1)\right)+i \sin \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(1)\right)\right]64^{\frac{1}{3}}=w_{1}
\]
Simplify:
\[
\begin{aligned}
=\left[\cos \left(30^{\circ}+120^{\circ}\right)+i \sin \left(30^{\circ}+120^{\circ}\right)\right]4 \\
=&\left[4\left[\cos 150^{\circ}+i \sin 150^{\circ}\right]\right.
\end{aligned}
\]
When $2=k$ :
\[
\left[\cos \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(2)\right)+i \sin \left(\frac{90^{\circ}}{3}+\frac{360^{\circ}}{3}(2)\right)\right]64^{\frac{1}{3}}=w_{2}
\]
Simplify:
\[
\begin{aligned}
=\left[\cos \left(30^{\circ}+240^{\circ}\right)+i \sin \left(30^{\circ}+240^{\circ}\right)\right]4 & \\
&=\left[\cos 270^{\circ}+i \sin 270^{\circ}\right]4
\end{aligned}
\]