How can you find the cubed roots of $i$? I am trying to figure out what the three possibilities of $z$ are such that 
$$
z^3=i
$$
but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.

As $\displaystyle i^2=-1,i=-i^3\implies z^3=i=-i^3=(-i)^3\iff z^3-(-i)^3=0$
Now, $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2),$
$\displaystyle\implies\{z-(-i)\}\{z^2+z(-i)+(-i)^2\}=\{z-(-i)\}(z^2-iz-1)$
If $z-(-i)=0, z=-i$
Else $z^2-iz-1=0\implies z=\dfrac{i\pm\sqrt{i^2-4(-1)}}2=\dfrac{i\pm\sqrt3}2$
