Can someone help me solve $z^{3}=-i$ $z$ is a complex number. I tried to solve this question by setting $z=a+bi$, but when I calculated $(a+bi)^{3}$, I found that's a little bit complicated to compute, Can someone help me teach me some easier way to solve the problem? Thanks a lot.

There was no need to give up on the $(a+bi)^3$ approach.  Just follow the algebra:
$$-i=(a+bi)^3=a^3+3a^2bi-3ab^2-b^3i=(a^3-3ab^2)+(3a^2b-b^3)i$$
implies
$$
a^3-3ab^2=0\quad\text{and}\quad
3a^2b-b^3=-1$$
Factoring the first equation as $a(a^2-3b^2)=0$, we have $a=0$ or $a=\pm\sqrt3b$.  
If $a=0$, then the second equation is simply $-b^3=-1$, for which $b=1$ is the only solution.
If $a=\pm3b$, the second equation is $9b^3-b^3=-1$, or $8b^3=-1$, for which $b=-1/2$ is the only solution.  We thus get three solutions to $z^3=-i$:
$$z=
\begin{cases}
i\\
{\sqrt3\over2}-{1\over2}\\
-{\sqrt3\over2}-{1\over2}
\end{cases}$$
