### Instruction: Find the solutions to $z^3 = -8.$  Enter the solutions, separated by commas.### Response: I notice that this is a cubic equation, and I wonder if I can factor it into a product of linear and quadratic terms.
I recall that a cubic equation always has at least one real root, so I look for a possible value of $z$ that would make $z^3 = -8.$
I see that $z = -2$ is such a value, since $(-2)^3 = -8.$
I use synthetic division to divide $z^3 + 8$ by $z + 2.$
I get $z^2 - 2z + 4.$
I check that this quotient is indeed a factor of the original equation, by multiplying it by $z + 2$ and verifying that I get $z^3 + 8.$
I now have $z^3 + 8 = (z + 2)(z^2 - 2z + 4),$ and I want to find the other solutions by setting the quadratic factor equal to zero.
I use the quadratic formula to find the roots of $z^2 - 2z + 4 = 0.$
I get $z = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm \sqrt{3}i,$ where $i$ is the imaginary unit.
I conclude that the three solutions to the original equation are $z = -2, 1 + \sqrt{3}i, 1 - \sqrt{3}i.$