+9519 Lemma: If \(x + \dfrac1x = 2\cos \alpha\), then \(x^n + \dfrac1{x^n} = 2\cos n\alpha\)
If you want to see the proof of this, go here. I have done the proof in a previous answer before.
Now, if we consider \(\alpha = \dfrac{2\pi}3\) and \(n = 99\), using the formula directly gives
\(x^{99} + \dfrac1{x^{99}} = 2\cos\left(\dfrac{2\pi}3\cdot 99\right) = \boxed{2}\)
