Explanation

(d)
Given, $x^{2} - \sqrt{3}x + 1 = 0$
$\Rightarrow \ \ \ x = \frac{\sqrt{3} \pm \sqrt{3 - 4}}{2} = \frac{\sqrt{3} \pm i}{2} = \cos{\frac{\pi}{6} \pm i\sin\frac{\pi}{6}}$
$\ \Rightarrow \ \ \ x^{n} = \cos{\frac{n\pi}{6} \pm i\sin\frac{n\pi}{6}}$
And $\frac{1}{x^{n}} = \cos{\frac{n\pi}{6} \pm i\sin\frac{n\pi}{6}}$
$\therefore\ \ \ x^{n} - \frac{1}{xn} = \pm 2i\sin\frac{n\pi}{6}$
$\Rightarrow \ \ \ \left( x^{n} - \frac{1}{x^{n}} \right)^{2} = - 4\sin^{2}\frac{n\pi}{6}$
Hence, $\sum_{n = 1}^{24}\left( x^{n} - \frac{1}{x^{n}} \right)^{2}$
$= - 4\left\lbrack \sin^{2}{\frac{\pi}{6} + \sin^{2}{\frac{2\pi}{6} + ... + \sin^{2}\frac{24\pi}{6}}} \right\rbrack$
$= - 4(12) = - 48$