+817 Find the solutions to \(z^3 = -8.\) Enter the solutions, separated by commas.
+845 if the question is find z because im not sure,
z^3 = -8
(-2)^3 = -8
therefore z = -2
+118687 That is correct Zeet. 
The only real solution is z=-2
but I think mathtoo wants the complex solutions too. It is a cubic so there are 3 solutions altogether.
There will be equally spaced at \(\frac{2\pi}{3}\) radians apart on the complex plane. So the others are at \(\frac{\pi}{3}\;\;and \;\;\frac{-\pi}{3}\)
The modulus or absolute value is 2
so
\(z=-2,\;\;2e^{(\pi/3)i},\;\;2e^{(-\pi/3)i}\)
\(2e^{(\frac{\pi}{3})i}\\=2[cos(\pi/3)+isin(\pi/3)]\\ =2[\frac{1}{2}+i\frac{\sqrt3}{2}]\\ =1+i\sqrt3\\~\\ 2e^{(\frac{\pi}{3})i}\\ =1-i\sqrt3\)
so
\(z=-2,\;\;1+i\sqrt3,\;\;1-i\sqrt3\)
Here is the diagramatic representation. Copied from Wolfram|Alpha.
