Answer: The roots are
\(z=6\left(\cos \frac{5\pi}{18}+\mathrm{i}\sin \frac{5\pi}{18}\right),\quad 6\left(\cos \frac{17\pi}{18}+\mathrm{i}\sin \frac{17\pi}{18}\right),\quad 6\left(\cos \frac{29\pi}{18}+\mathrm{i}\sin \frac{29\pi}{18}\right).\)
Write
\(z^3=-108\sqrt{3}+108\mathrm{i}\)
in modulus-argument form.
The modulus is
\(|z^3|=\sqrt{(-108\sqrt{3})^2+108^2}=\sqrt{34992+11664}=\sqrt{46656}=216.\)
For the argument,
\(\tan\theta=\dfrac{108}{-108\sqrt{3}}=-\dfrac{1}{\sqrt{3}}.\)
Since the point \((-108\sqrt{3},108)\) is in the second quadrant,
\(\theta=\frac{5\pi}{6}.\)
So
\(z^3=216\left(\cos\frac{5\pi}{6}+\mathrm{i}\sin\frac{5\pi}{6}\right).\)
Taking cube roots, the modulus of each root is
\(\sqrt[3]{216}=6,\)
and the arguments are
\(\frac{\frac{5\pi}{6}+2k\pi}{3}\) for \(k=0,1,2\).
Hence
for \(k=0\): \(\frac{5\pi}{18}\),
for \(k=1\): \(\frac{5\pi/6+2\pi}{3}=\frac{17\pi}{18}\),
for \(k=2\): \(\frac{5\pi/6+4\pi}{3}=\frac{29\pi}{18}.\)
Therefore the three roots are
\(z_1=6\left(\cos \frac{5\pi}{18}+\mathrm{i}\sin \frac{5\pi}{18}\right),\)
\(z_2=6\left(\cos \frac{17\pi}{18}+\mathrm{i}\sin \frac{17\pi}{18}\right),\)
\(z_3=6\left(\cos \frac{29\pi}{18}+\mathrm{i}\sin \frac{29\pi}{18}\right).\)
