### Instruction: Find the solutions to $z^3 = -8.$  Enter the solutions, separated by commas.### Response: I know that one solution is $z = -2$, because $(-2)^3 = -8$.
To find the other solutions, I can divide both sides of the equation by $z + 2$, using polynomial long division or synthetic division.
I get $(z^3 + 8) / (z + 2) = z^2 - 2z + 4$.
Now I need to solve the quadratic equation $z^2 - 2z + 4 = 0$.
I can use the quadratic formula to find the solutions: $z = (-b \pm \sqrt{b^2 - 4ac}) / 2a$, where $a = 1, b = -2, c = 4$.
Plugging in these values, I get $z = (2 \pm \sqrt{4 - 16}) / 2$.
Simplifying, I get $z = (2 \pm \sqrt{-12}) / 2$.
Since the square root of a negative number is not a real number, I need to use imaginary numbers to express the solutions.
I can write $\sqrt{-12}$ as $\sqrt{-1} \sqrt{12}$, and use the fact that $\sqrt{-1} = i$, the imaginary unit.
I can also simplify $\sqrt{12}$ by factoring out the largest perfect square: $\sqrt{12} = \sqrt{4} \sqrt{3} = 2 \sqrt{3}$.
So, the solutions are $z = (2 \pm 2 \sqrt{3} i) / 2$.
I can simplify this further by dividing both the numerator and the denominator by 2: $z = 1 \pm \sqrt{3} i$.