Answer
The complex cube root of the given complex numbers in polar form are,
$2\left[ \cos 310{}^\circ +i\sin 310{}^\circ \right],\text{ }2\left[ \cos 70{}^\circ +i\sin 70{}^\circ \right],\text{ }2\left[ \cos 190{}^\circ +i\sin \cos 190{}^\circ \right]$.
Work Step by Step
Consider the given complex number,
$8\left( \cos 210{}^\circ +i\sin 210{}^\circ \right)$
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, using (1) to get,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ }{n} \right)+i\sin \left( \frac{\theta +360{}^\circ }{n} \right) \right] \\
& {{z}_{1}}=\sqrt[3]{8}\left[ \cos \left( \frac{210{}^\circ +360{}^\circ }{3} \right)+i\sin \left( \frac{210{}^\circ +360{}^\circ }{3} \right) \right] \\
& =2\left[ \cos \left( \frac{570{}^\circ }{3} \right)+i\sin \left( \frac{570{}^\circ }{3} \right) \right] \\
& =2\left[ \cos 190{}^\circ +i\sin 190{}^\circ \right]
\end{align}$
Similarly, for the other root,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ k}{n} \right)+i\sin \left( \frac{\theta +360{}^\circ k}{n} \right) \right] \\
& {{z}_{0}}=\sqrt[3]{8}\left[ \cos \left( \frac{210{}^\circ +360{}^\circ \times 0}{3} \right)+i\sin \left( \frac{210{}^\circ +360{}^\circ \times 0}{3} \right) \right] \\
& =2\left[ \cos \frac{210{}^\circ }{3}+i\sin \frac{210{}^\circ }{3} \right] \\
& =2\left[ \cos 70{}^\circ +i\sin 70{}^\circ \right]
\end{align}$
Similarly, for the other root,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +360{}^\circ k}{n} \right)+i\sin \left( \frac{\theta +360{}^\circ k}{n} \right) \right] \\
& {{z}_{2}}=\sqrt[3]{8}\left[ \cos \left( \frac{210{}^\circ +360{}^\circ \times 2}{3} \right)+i\sin \left( \frac{210{}^\circ +360{}^\circ \times 2}{3} \right) \right] \\
& =2\left[ \cos \frac{930{}^\circ }{3}+i\sin \frac{930{}^\circ }{3} \right] \\
& =2\left[ \cos 310{}^\circ +i\sin 310{}^\circ \right]
\end{align}$
Therefore,
The complex cube root of the given complex numbers in polar form are,
$2\left[ \cos 310{}^\circ +i\sin 310{}^\circ \right],\text{ }2\left[ \cos 70{}^\circ +i\sin 70{}^\circ \right],\text{ }2\left[ \cos 190{}^\circ +i\sin \cos 190{}^\circ \right]$.
