I understand that the way to calculate the cube root of $i$ is to use Euler's formula and divide $\frac{\pi}{2}$ by $3$ and find $\frac{\pi}{6}$ on the complex plane; however, my question is why the following solution doesn't work.
So $(-i)^3 = i$, but why can I not cube root both sides and get $-i=(i)^{\frac{1}{3}}$. Is there a rule where equality is not maintained when you root complex numbers or is there something else I am violating and not realizing?
On
$-i$ is certainly one of the cube roots of $i$ but it is certainly not the only one.
Every complex number (except $0$) has three cube roots.
A quicker way to find these roots is to use the cube roots of unity, which can be written $1, \omega, \omega^2$ and multiply them successively by the root you've already got.
So in your case, the three roots are $-i, - \omega i =\frac{\sqrt3}{2} + \frac 12 i, - \omega^2 i = -\frac{\sqrt3}{2} + \frac 12 i$
There are three cube roots of $i$. The value at $e^{i\pi/6}$ is simply one of the roots. To find all of the roots, you can add $2\pi/3$ for each root to the angle of $\pi/6$. Since one root is at $\pi/6$, the next one will be at $\pi/6 + 2\pi/3 = 5\pi/6$. The next one will be at $5\pi/6 + 2\pi/3 = 3\pi/2$. With this last angle, the root is exactly $-i$.
In other words, given one of the roots at angle $\theta_0$, all of them are $e^{i(\theta_0+2n\pi/3)}$, where $n$ goes through $0, 1, 2$.