Answer
The complex cube roots of the given complex numbers in polar form are,
$3\left[ \cos 342{}^\circ +i\sin 342{}^\circ \right]\text{, }3\left[ \cos 102{}^\circ +i\sin 102{}^\circ \right]\text{ and }3\left[ \cos 222{}^\circ +i\sin 222{}^\circ \right]$.
Work Step by Step
Consider the given complex number,
$27\left( \cos 306{}^\circ +i\sin 306{}^\circ \right)$
For any complex number of the type,
$z=r\cos \theta +i\sin \theta $
If n is a positive integer, then ${{z}^{n}}$ is,
${{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ k}{n} \right)+i\sin \left( \frac{\theta +360{}^\circ k}{n} \right) \right]$ ...... (1)
Now, for the given complex number, use (1) to get,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ k}{n} \right)+i\sin \left( \frac{\theta +360{}^\circ k}{n} \right) \right] \\
& {{z}_{1}}=\sqrt[3]{27}\left[ \cos \left( \frac{306{}^\circ +360{}^\circ }{3} \right)+i\sin \left( \frac{306{}^\circ +360{}^\circ }{3} \right) \right] \\
& {{z}_{1}}=3\left[ \cos \left( \frac{666{}^\circ }{3} \right)+i\sin \left( \frac{666{}^\circ }{3} \right) \right] \\
\end{align}$ \
Now, for the other root,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ k}{n} \right)+i\sin \left( \frac{\theta +360{}^\circ k}{n} \right) \right] \\
& {{z}_{0}}=\sqrt[3]{27}\left[ \cos \left( \frac{306{}^\circ +360{}^\circ \times 0}{3} \right)+i\sin \left( \frac{306{}^\circ +360{}^\circ \times 0}{3} \right) \right] \\
& {{z}_{0}}=3\left[ \cos \frac{306{}^\circ }{3}+i\sin \frac{306{}^\circ }{3} \right] \\
& {{z}_{0}}=3\left[ \cos 102{}^\circ +i\sin 102{}^\circ \right] \\
\end{align}$ \
Now, for the other root,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ }{n} \right)+i\sin \left( \frac{\theta +360{}^\circ }{n} \right) \right] \\
& {{z}_{2}}=\sqrt[3]{27}\left[ \cos \left( \frac{306{}^\circ +360{}^\circ \times 2}{3} \right)+i\sin \left( \frac{306{}^\circ +360{}^\circ \times 2}{3} \right) \right] \\
& {{z}_{2}}=3\left[ \cos \frac{1026{}^\circ }{3}+i\sin \frac{1026{}^\circ }{3} \right] \\
& {{z}_{2}}=3\left[ \cos 342{}^\circ +i\sin 342{}^\circ \right] \\
\end{align}$
The complex cube roots of the given complex numbers in polar form are,
$3\left[ \cos 342{}^\circ +i\sin 342{}^\circ \right]\text{, }3\left[ \cos 102{}^\circ +i\sin 102{}^\circ \right]\text{ and }3\left[ \cos 222{}^\circ +i\sin 222{}^\circ \right]$.
