Ther are 3 roots.
The first one is easy to find it is 2
There are 2pi radians in a circle and we want 3 roots so each will be 2pi/3 radians apart.
The roots will be
2, 2 cis(2pi/3) and 2cis(-2pi/3)
\(cos(\frac{2\pi}{3}) = - cos (\frac{\pi}{3}) =\frac{ -1}{2}\\ sin(\frac{2\pi}{3})= sin(\frac{\pi}{3}) = \frac{\sqrt3}{2}\\ cis(\frac{2\pi}{3}) =\frac{ -1}{2}+i*\frac{\sqrt3}{2}\\ cis(\frac{2\pi}{3}) =\frac{ -1+\sqrt{3}\;i}{2}\\~\\ cos(\frac{-2\pi}{3}) = - cos (\frac{\pi}{3}) =\frac{ -1}{2}\\ sin(\frac{-2\pi}{3})= -sin(\frac{\pi}{3}) = \frac{-\sqrt3}{2}\\ cis(\frac{-2\pi}{3}) =\frac{ -1}{2}-i*\frac{\sqrt3}{2}\\ cis(\frac{-2\pi}{3}) =\frac{ -1-\sqrt{3}\;i}{2}\\ \)
\(\mbox{So the 3 solutions are }\\~\\ \sqrt[3]{8}=2,\qquad and \quad\\ \sqrt[3]{8} = -1+\sqrt{3}\;i\quad and \quad\\ \sqrt[3]{8} = -1-\sqrt{3}\;i, \\\)
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