### Instruction: Find the solutions to $z^3 = -8.$  Enter the solutions, separated by commas.### Response: To solve this equation, I need to find the complex roots of -8.
I can write -8 in polar form as $-8 = 8(\cos \pi + i \sin \pi)$, using Euler's formula.
Then, using the formula for the nth roots of a complex number, I can find the three cube roots of -8 as $z_k = 2(\cos (\frac{\pi + 2k\pi}{3}) + i \sin (\frac{\pi + 2k\pi}{3}))$, for $k = 0, 1, 2$.
Simplifying, I get $z_0 = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 1 + i \sqrt{3}$, $z_1 = 2(\cos \pi + i \sin \pi) = -2$, and $z_2 = 2(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3}) = 1 - i \sqrt{3}$.
These are the three solutions to the equation.