Answer: The roots are

\(z=14^{1/3}\mathrm{e}^{-\mathrm{i}\pi/18},\quad 14^{1/3}\mathrm{e}^{\mathrm{i}11\pi/18},\quad 14^{1/3}\mathrm{e}^{-13\mathrm{i}\pi/18}.\)

Write \(7\sqrt{3}-7\mathrm{i}\) in modulus-argument form.

Its modulus is

\(|7\sqrt{3}-7\mathrm{i}|=\sqrt{(7\sqrt{3})^2+(-7)^2}=\sqrt{147+49}=14.\)

Its argument is given by

\(\tan\theta=\dfrac{-7}{7\sqrt{3}}=-\dfrac{1}{\sqrt{3}}.\)

Since the number lies in the fourth quadrant, \(\theta=-\dfrac{\pi}{6}\).

So

\(z^3=14\mathrm{e}^{-\mathrm{i}\pi/6}.\)

Let \(z=re^{\mathrm{i}\phi}\). Then

\(r^3=14\Rightarrow r=14^{1/3},\)

and

\(3\phi=-\dfrac{\pi}{6}+2k\pi,\qquad k=0,1,2.\)

Hence

\(\phi=-\dfrac{\pi}{18}+\dfrac{2k\pi}{3}.\)

Taking \(k=0,1,2\):

• \(k=0:\ \phi=-\dfrac{\pi}{18}\)
• \(k=1:\ \phi=-\dfrac{\pi}{18}+\dfrac{2\pi}{3}=\dfrac{11\pi}{18}\)
• \(k=2:\ \phi=-\dfrac{\pi}{18}+\dfrac{4\pi}{3}=\dfrac{23\pi}{18}=-\dfrac{13\pi}{18}\) in the required range

Therefore the roots, with \(-\pi\leqslant \theta\lt \pi\), are

\(\boxed{14^{1/3}\mathrm{e}^{-\mathrm{i}\pi/18},\quad 14^{1/3}\mathrm{e}^{\mathrm{i}11\pi/18},\quad 14^{1/3}\mathrm{e}^{-13\mathrm{i}\pi/18}.}\)