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.

You can solve this geometrically if you know polar coordinates.
In polar coordinates, multiplication goes $(r_1, \theta_1) \cdot (r_2, \theta_2) = (r_1 \cdot r_2, \theta_1 + \theta_2)$, so cubing goes $(r, \theta)^3 = (r^3, 3\theta)$. The cube roots of $(r, \theta)$ are $\left(\sqrt[3]{r}, \frac{\theta}{3}\right)$, $\left(\sqrt[3]{r}, \frac{\theta+2\pi}{3}\right)$ and $\left(\sqrt[3]{r}, \frac{\theta+4\pi}{3}\right)$ (recall that adding $2\pi$ to the argument doesn't change the number). In other words, to find the cubic roots of a complex number, take the cubic root of the absolute value (the radius) and divide the argument (the angle) by 3.
$i$ is at a right angle from $1$: $i = \left(1, \frac{\pi}{2}\right)$. Graphically:

A cubic root of $i$ is $A = \left(1, \frac{\pi}{6}\right)$. The other two are $B = \left(1, \frac{5\pi}{6}\right)$ and $\left(1, \frac{9\pi}{6}\right) = -i$.
Recalling basic trigonometry, the rectangular coordinates of $A$ are $\left(\cos\frac{\pi}{6}, \sin\frac{\pi}{6}\right)$ (the triangle OMA is rectangle at M). Thus, $A = \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + i\frac{1}{2}$.
If you don't remember the values of $\cos\frac{\pi}{6}$ and $\sin\frac{\pi}{6}$, you can find them using geometry. The triangle $OAi$ has two equal sides $OA$ and $Oi$, so it is isoceles: the angles $OiA$ and $OAi$ are equal. The sum of the angles of the triangle is $\pi$, and we know that the third angle $iOA$ is $\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$; therefore $OiA = OAi = \dfrac{\pi - \frac{pi}{3}}{2} = \dfrac{\pi}{3}$. So $OAi$ is an equilateral triangle, and the altitude AN is also a median, so N is the midpoint of $[Oi]$: $\sin\frac{\pi}{6} = AM = ON = \frac{1}{2}$. By the Pythagorean theorem, $OM^2 + AM^2 = OA^2 = 1$ so $\cos\frac{\pi}{6} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \dfrac{\sqrt{3}}{2}$.
