Expressing $z^3 = -i$ in the form $z = a +bi$ How would I go about expressing $z^3 = -i$ in the form $z = a + bi$ without any trig functions in the answer?

I know that $z = a + ib$ so would I go ahead and solve that equaling $-i$?
Hint: use that 

$$z^3+i = z^3-i^3=(z-i)(z^2+zi+i^2)$$

So $z_1=i$ and the other two you will get solving $z^2+zi-1=0$.

That you can do either by direct solving a quadratic equation or plug in $z=a+bi$ with $a,b$ real...
For $z = a+ib$, we have
\begin{equation}
 z^3 = (a+ib)^3 = (\underbrace{a^3 - 3ab^2}_0 )+(\underbrace{3a^2b- b^3}_{-1})i
\end{equation}
Now solve for $a,b$ using the two equations above and you must get three distinct solutions $(a_1,b_1), (a_2,b_2), (a_3,b_3)
$