### Instruction: Find the solutions to $z^3 = -8.$  Enter the solutions, separated by commas.### Response: I recognize that $-8$ is a cube of $-2$, so one solution is $z = -2$.
However, I also know that complex numbers can have more than one cube root, so I need to find the others.
I recall that the cube roots of unity are $1, e^{2\pi i/3}$, and $e^{4\pi i/3}$, where $i$ is the imaginary unit.
So, if I multiply $-2$ by each of these cube roots of unity, I should get the other solutions.
I get $-2, -2e^{2\pi i/3}$, and $-2e^{4\pi i/3}$.
To write these in standard complex form, I need to use the fact that $e^{i\theta} = \cos \theta + i \sin \theta$.
I get $-2, -2(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}), -2(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3})$.
Simplifying the trigonometric functions, I get $-2, -2(-\frac{1}{2} + i \frac{\sqrt{3}}{2}), -2(-\frac{1}{2} - i \frac{\sqrt{3}}{2})$.
Multiplying by $-2$, I get $-2, 1 - i \sqrt{3}, 1 + i \sqrt{3}$.
These are the three solutions to the equation.