<s>[INST] $x^{2000} + \frac{1}{x^{2000}}$ in terms of $x + \frac 1x$. If $x + \frac{1}{x} = 1$, then what is
$$ x^{2000} + \frac{1}{x^{2000}} = ?$$
 [/INST] Assume that $x=e^{i\theta}$. Then you are trying to write $2\cos(2000\cdot\theta)$ in terms of $2\cos(\theta)$, so the answer is given by Chebyshev polynomials of the first kind:
$$ \left(x^{2000}+\frac{1}{x^{2000}}\right) = 2\cdot \widetilde{T}_{\!2000}\left(x+\frac{1}{x}\right) $$
where $\widetilde{T}_{n}(z) = T_n(z/2)$. In our very particular case, we may notice that $\frac{1}{x}+x=1$ implies $x=e^{\pm \pi i/3}$: since $2000\equiv 2\pmod{6}$, given $x+\frac{1}{x}=1$ we have:
$$ x^{2000}+\frac{1}{x^{2000}}=x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 = \color{red}{-1}.$$
</s>