Answer: The roots are
\(z=3\times 2^{1/6}e^{-i\pi/12},\quad 3\times 2^{1/6}e^{i7\pi/12},\quad 3\times 2^{1/6}e^{-i3\pi/4}.\)
First write \(27-27i\) in modulus-argument form.
Its modulus is
\(r=\sqrt{27^2+(-27)^2}=27\sqrt{2}.\)
Its argument is \(-\pi/4\), since the point \((27,-27)\) lies in the fourth quadrant.
So
\(27-27i=27\sqrt{2}\,e^{-i\pi/4}.\)
If \(z^3=27\sqrt{2}\,e^{-i\pi/4}\), then the cube roots are
\(z=(27\sqrt{2})^{1/3}e^{i\left(\frac{-\pi/4+2k\pi}{3}\right)}\), where \(k=0,1,2\).
Now
\((27\sqrt{2})^{1/3}=27^{1/3}(\sqrt{2})^{1/3}=3\times 2^{1/6}.\)
So
\(z=3\times 2^{1/6}e^{i\left(\frac{-\pi/4+2k\pi}{3}\right)}.\)
Taking the three values of \(k\):
For \(k=0\): \(\theta=\frac{-\pi/4}{3}=-\pi/12\), so \(z=3\times 2^{1/6}e^{-i\pi/12}.\)
For \(k=1\): \(\theta=\frac{-\pi/4+2\pi}{3}=\frac{7\pi}{12}\), so \(z=3\times 2^{1/6}e^{i7\pi/12}.\)
For \(k=2\): \(\theta=\frac{-\pi/4+4\pi}{3}=\frac{15\pi}{12}=\frac{5\pi}{4}.\) This is outside the required range \(-\pi\leq \theta<\pi\), so subtract \(2\pi\): \(\frac{5\pi}{4}-2\pi=-\frac{3\pi}{4}.\) Hence \(z=3\times 2^{1/6}e^{-i3\pi/4}.\)
Therefore the three roots are
\(\boxed{3\times 2^{1/6}e^{-i\pi/12},\quad 3\times 2^{1/6}e^{i7\pi/12},\quad 3\times 2^{1/6}e^{-i3\pi/4}}.\)
