I solved it 
\(tan(\theta)=\frac{y}{x}=\frac{b}{a}=\frac{4}{4\sqrt{3}}\)
\(\theta=\frac{\pi}{6}\)
\(r=8\)
In polar form
\(8[cos(\frac{\pi}{6})+isin(\frac{\pi}{6})] \)
Now to find the roots.
\(\sqrt[n]{8}[cos(\frac{\frac{\pi}{6}+2\pi k}{n})+isin(\frac{\frac{\pi}{6}+2\pi k}{n})]\)
This question asks for cube roots so n=3
\(\sqrt[3]{8}[cos(\frac{\frac{\pi}{6}+2\pi k}{3})+isin(\frac{\frac{\pi}{6}+2\pi k}{3})]\)
Now that this is all filled in, lets change the k to each root we need.
\(k=0: 2[cos(\frac{\frac{\pi}{6}}{3})+isin(\frac{\frac{\pi}{6}}{3})]\)
\(k=1: 2[cos(\frac{\frac{\pi}{6}+2\pi}{3})+isin(\frac{\frac{\pi}{6}+2\pi}{3})]\)
\(k=2: 2[cos(\frac{\frac{\pi}{6}+4\pi }{3})+isin(\frac{\frac{\pi}{6}+4\pi }{3})]\)
.
AGAIN,SORRY
asinus
