TITLE: How do I find $\sqrt[3]{-i}$?
QUESTION [5 upvotes]: I'm asked to evaluate $$\sqrt[3]{-i}$$
I suppose $\sqrt[3]{-i}=(a+bi)$
$$\implies (a+bi)^3=-i$$
$$\implies \Im \left( (a+bi)^3 \right) =\Im \left( (-i) \right)$$
$$\implies 3a^2b-b^3=-1$$
Now how am I supposed to find $a$,$b$? 
Aren't there infinitely of them instead of just three?

REPLY [2 votes]: Alt. hint:   the cube roots of $\,-i\,$ are the solutions to $\,z^3=-i \iff z^3+i=0\,$. Using that $\,i = -i^3\,$ and the identity $\,a^3-b^3=(a-b)(a^2+ab+b^2)\,$, the latter equation can be written as:
$$
0 = z^3+i = z^3-i^3=(z-i)(z^2+iz + i^2)=(z-i)\big(-(iz)^2+iz-1\big)
$$
The first factor gives the (obvious) solution $\,z=i\,$, and the second factor is a real quadratic in $\,iz\,$ with roots $\,iz = (1 \pm i \sqrt{3})/2 \iff z = \ldots$