so... let's see, so is -8, or -8 + 0i then
[tex]\bf \begin{array}{rllll}
-8&,&0i\\
a&&b
\end{array}\qquad r=\sqrt{a^2+b^2}\implies r=\sqrt{(-8)^2+0^2}\implies r=8
\\\\\\
\textit{and notice the picture below}\qquad \theta=\pi [/tex]
[tex]\bf \textit{ roots of complex numbers, DeMoivre's theorem}
\\\\
\sqrt[{{ n}}]{z}=\sqrt[{{ n}}]{r}\left[ cos\left( \cfrac{\theta+2\pi k}{{{ n}}} \right) +i\ sin\left( \cfrac{\theta+2\pi k}{{{ n}}} \right)\right]\quad
\begin{array}{llll}
k\ roots\\
0,1,2,3,...
\end{array}\\\\
-------------------------------\\\\
z=-8+0i\qquad \qquad \sqrt[3]{z}=\sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi k}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi k}{{{ 3}}} \right)\right]\\\\[/tex]
[tex]\bf -------------------------------\\\\
\sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi (0)}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi (0)}{{{ 3}}} \right)\right]\impliedby
\begin{array}{llll}
first\ root\\
k=0
\end{array}
\\\\\\
2\left[ cos\left( \frac{\pi }{3} \right)+i\ sin\left( \frac{\pi }{3} \right)\right] \implies 2\cdot \cfrac{1}{2}+2\cdot \cfrac{\sqrt{3}}{2}\ i\implies 1+\sqrt{3}\ i[/tex]
[tex]\bf -------------------------------\\\\
\sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi (1)}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi (1)}{{{ 3}}} \right)\right]\impliedby
\begin{array}{llll}
second\ root\\
k=1
\end{array}
\\\\\\
2\left[ cos\left( \pi \right)+i\ sin\left( \pi \right)\right]\implies 2\cdot -1+2\cdot 0\ i\implies -2+0i[/tex]
[tex]\bf -------------------------------\\\\
\sqrt[{{ 3}}]{8}\left[ cos\left( \cfrac{\pi+2\pi (2)}{{{ 3}}} \right) +i\ sin\left( \cfrac{\pi+2\pi (2)}{{{ 3}}} \right)\right]\impliedby
\begin{array}{llll}
third\ root\\
k=2
\end{array}
\\\\\\
2\left[ cos\left( \frac{5\pi }{3} \right)+i\ sin\left( \frac{5\pi }{3} \right)\right]\implies 2\cdot \cfrac{1}{2}+2\cdot -\cfrac{\sqrt{3}}{2}\ i\implies 1-\sqrt{3}\ i[/tex]
