What are the three cube roots of -1? What are the three cube roots of -1?
Not sure if this is a trick question, But I have been asked this. one of the ansers is -1, what are the other 2?

Write $-1$ in polar form as $e^{i\pi}$. In general, the cube roots of $r e^{i\theta}$ are given by $r^{1/3}e^{i\theta/3}$, $r^{1/3}e^{i(\theta/3 + 2\pi /3)}$ and $r^{1/3}e^{i(\theta/3 + 4\pi /3)}$. In your case $r = 1$ and $\theta = \pi$, so your cube roots are $e^{i\pi / 3}$, $e^{i\pi}$, and $e^{i 5\pi/ 3}$. Put back into rectangular form, they are ${1 \over 2} + i{\sqrt{3} \over 2}$, $-1$, and ${1 \over 2} - i{\sqrt{3} \over 2}$.
