Answer
The three cube roots are:
\[
\begin{array}{l}
[\cos 101+i \sin 101] 3\\
[\cos 221+i \sin 221] 3\\
[\cos 341+i \sin 341]3
\end{array}
\]
Work Step by Step
\[
\left(\cos 303^{\circ}+i \sin 303^{\circ}\right)27
\]
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 $0=k$ :
\[
\left[\cos \left(\frac{303^{\circ}}{3}+\frac{360^{\circ}}{3}(0)\right)+i \sin \left(\frac{303^{\circ}}{3}+\frac{360^{\circ}}{3}(0)\right)\right]27^{\frac{1}{3}}=w_{0}
\]
Simplify:
\[
=\sqrt{3 \cos 101^{\circ}+i \sin 101^{\circ}}
\]
When $1=k$ :
\[
\left[\cos \left(\frac{303^{\circ}}{3}+\frac{360^{\circ}}{3}(1)\right)+i \sin \left(\frac{303^{\circ}}{3}+\frac{360^{\circ}}{3}(1)\right)\right]27^{\frac{1}{3}}=w_{1}
\]
Simplify:
\[
\begin{array}{r}
=\left[\cos \left(101^{\circ}+120^{\circ}\right)+i \sin \left(101^{\circ}+120^{\circ}\right)\right]3 \\
=\left[3\left[\cos 221^{\circ}+i \sin 221^{\circ}\right]\right.
\end{array}
\]
When $2=k$ :
\[
\left[\cos \left(\frac{303^{\circ}}{3}+\frac{360^{\circ}}{3}(2)\right)+i \sin \left(\frac{303^{\circ}}{3}+\frac{360^{\circ}}{3}(2)\right)\right]27^{\frac{1}{3}}=w_{2}
\]
Simplify:
\[
\begin{array}{r}
=\left[\cos \left(101^{\circ}+240^{\circ}\right)+i \sin \left(101^{\circ}+240^{\circ}\right)\right] \\
=\left[\cos 341^{\circ}+i \sin 341^{\circ}\right]3
\end{array}
\]
