jmc

From the equation $z^3=-8,$ $z^3+8=0.$ We see that $z=-2$ is one solution, so we can take out a factor of $z+2=0,$ which gives us $$(z+2)(z^2-2z+4)=0.$$By the quadratic formula, the roots of $z^2-2z+4=0$ are $$z=\frac{2\pm \sqrt{2^2-4\cdot 4}}{2}=\frac{2\pm 2i\sqrt{3}}{2}=1\pm i\sqrt{3}.$$Thus, the solutions are $\boxed{-2,1+i\sqrt{3},1-i\sqrt{3}}.$