Answer
See the answer below
Work Step by Step
We need to show that
$(\frac{1}{2}+\frac{\sqrt 3}{2}i)^{3}=-1$
$\frac{1}{2}+\frac{\sqrt 3}{2}i$ in trigonometric form is
$1(\cos60^{\circ}+i\sin60^{\circ})$
Then, using de Moivre's theorem, we get
$(\frac{1}{2}+\frac{\sqrt 3}{2}i)^{3}=[1(\cos60^{\circ}+i\sin60^{\circ})]^{3}$
$=(1)^{3}[\cos (3\cdot60^{\circ})+i\sin (3\cdot60^{\circ})]$
$=(\cos 180^{\circ}+i\sin180^{\circ})$
which in standard form is
$=-1+i\cdot0=-1$
Hence we proved that $\frac{1}{2}+\frac{\sqrt 3}{2}i$ is a cube root of -1.
