### Instruction: Find the solutions to $z^3 = -8.$  Enter the solutions, separated by commas.### Response: To solve this problem, I need to find the complex roots of -8.
I recall that any complex number can be written in polar form as $r e^{i \theta}$, where $r$ is the modulus and $\theta$ is the argument of the complex number.
So, I can rewrite $z^3 = -8$ as $z^3 = 8 e^{i \pi}$, since $-8$ has a modulus of 8 and an argument of $\pi$.
Now, I can use the fact that $z^3 = r^3 e^{3i \theta}$, where $r$ and $\theta$ are the modulus and argument of $z$, respectively.
This gives me the equation $r^3 e^{3i \theta} = 8 e^{i \pi}$.
To find $r$, I can compare the moduli of both sides and get $r^3 = 8$, which implies $r = 2$.
To find $\theta$, I can compare the arguments of both sides and get $3 \theta = \pi + 2 \pi k$, where $k$ is any integer.
This implies $\theta = \frac{\pi}{3} + \frac{2 \pi k}{3}$, where $k = 0, 1, 2$.
Therefore, the solutions for $z$ are $z = 2 e^{i \pi/3}$, $z = 2 e^{i \pi}$, and $z = 2 e^{5i \pi/3}$.
In rectangular form, these are $z = 2 \cos \frac{\pi}{3} + 2 i \sin \frac{\pi}{3}$, $z = -2$, and $z = 2 \cos \frac{5 \pi}{3} + 2 i \sin \frac{5 \pi}{3}$.
Simplifying, these are $z = 1 + i \sqrt{3}$, $z = -2$, and $z = 1 - i \sqrt{3}$.