$\displaystyle - i = \cos \left( {\frac{{ - \pi }}{2}} \right) + i\sin \left( {\frac{{ - \pi }}{2}} \right)$.
Now find the three cube roots. One of then is
$\displaystyle \cos \left( {\frac{{ - \pi }}{6}} \right) + i\sin \left( {\frac{{ - \pi }}{6}} \right)$.
What are the other two?
One of the cube roots is $\displaystyle \cos{\left(-\frac{\pi}{6}\right)} + i\sin{\left(-\frac{\pi}{6}\right)}$.
All of the cube roots are evenly spaced around the unit circle.
So they are all separated by an angle of $\displaystyle \frac{2\pi}{3}$.
The radius is 1, so the other roots are at
$\displaystyle -\frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{2}$
and
$\displaystyle \frac{\pi}{2} + \frac{2\pi}{3} = \frac{7\pi}{6} = -\frac{5\pi}{6}$.
Therefore the three roots are
$\displaystyle \textrm{cis}\,\left(-\frac{\pi}{6}\right), \textrm{cis}\,\left(\frac{\pi}{6}\right), \textrm{cis}\,\left(-\frac{5\pi}{6}\right)$.