A problem by milind prabhu

$$ Let $a=e^{i\theta}$ , then $a^{-1}=e^{-i\theta}$ . Euler's formula states that, for any real number $\theta$ and $n$ , $$ $e^{in\theta}=\cos n\theta+i\sin n\theta$ and $e^{-in\theta}=\cos n\theta-i\sin n\theta.$ $$ Then $\begin{aligned} a+a^{-1}&=\sqrt{3}\\ e^{i\theta}+e^{-i\theta}&=\sqrt{3}\\ \cos\theta+i\sin\theta+\cos\theta-i\sin\theta&=\sqrt{3}\\ 2\cos\theta&=\sqrt{3}\\ \theta&=\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\\ &=\frac{\pi}{6}\text{rad}\\ &=30^\circ. \end{aligned}$ $$ Therefore, for $n=3$ , we obtain $$ $\begin{aligned} a^3+a^{-3}&=e^{3i\theta}+e^{-3i\theta}\\ a^3+\frac{1}{a^3}&=2\cos(3\theta)\\ &=2\cos(3\cdot30^\circ)\\ &=2\cos(90^\circ)\\ &=2\cdot 0\\ &=\boxed{0} \end{aligned}$ $$