### Instruction: Find the solutions to $z^3 = -8.$  Enter the solutions, separated by commas.### Response: To find the solutions to $z^3 = -8,$ I need to find the complex numbers that satisfy this equation.
One way to approach this is to write $z$ in polar form, as $z = r(\cos\theta + i\sin\theta),$ where $r$ is the modulus and $\theta$ is the argument of $z.$
Then, using the identity $(r_1e^{i\theta_1})(r_2e^{i\theta_2}) = r_1r_2e^{i(\theta_1+\theta_2)},$ I can rewrite the equation as $r^3(\cos 3\theta + i\sin 3\theta) = 8(\cos\pi + i\sin\pi).$
This implies that $r^3 = 8$ and $3\theta = \pi + 2k\pi,$ where $k$ is any integer.
Solving for $r,$ I get $r = 2,$ since $r$ has to be positive and $2^3 = 8.$
Solving for $\theta,$ I get $\theta = \frac{\pi}{3} + \frac{2k\pi}{3},$ where $k = 0, 1,$ or $2.$
If $k > 2,$ the angle would be coterminal with one of the previous values, so I don't need to consider them.
Therefore, the three solutions for $z$ are
\begin{align*}
z_0 &= 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 1 + i\sqrt{3},\\
z_1 &= 2\left(\cos\pi + i\sin\pi\right) = 2\left(-1 + i\cdot 0\right) = -2,\\
z_2 &= 2\left(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right) = 2\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 1 - i\sqrt{3}.
\end{align*}