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If \(x + \frac{1}{x} = -1,\) find \(x^{99} + \frac{1}{x^{99}}\)

 Jul 8, 2020
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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

 

\(x^{99} + \dfrac1{x^{99}} = 2\cos\left(\dfrac{2\pi}3\cdot 99\right) = \boxed{2}\)

 Jul 8, 2020

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