$x^{2000} + \frac{1}{x^{2000}}$ in terms of $x + \frac 1x$. If $x + \frac{1}{x} = 1$, then what is
$$ x^{2000} + \frac{1}{x^{2000}} = ?$$
 A: 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:
$$ x^{2000}+\frac{1}{x^{2000}}=x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 = \color{red}{-1}.$$
A: $$x^2-x+1=0\implies x^3+1=(x+1)(x^2-x+1)=0\iff x^3=-1$$
$$x^{2000}=(x^3)^{666}\cdot x^2=(-1)^{666}\cdot x^2=x^2$$
Finally use  $a^2+b^2=(a+b)^2-2ab$ for $x^2+\dfrac1{x^2}$
A: This is a not so elegant approach. 
Let $a_n=x^n+\frac1{x^n}$ then 
$$a_n\left(x+\frac1x\right)=a_{n+1}+a_{n-1}$$ or $$a_{n+1}=a_{n}-a_{n-1}$$
We have $a_1=1$ and $a_2=-1$, hence solving recursively - plugging - we obtain \begin{align}
a_{6n-4}=a_{6n-2}&=-1\\
a_{6n-3}&=-2\\
a_{6n}&=2\\
a_{6n-1}=a_{6n+1}&=1
\end{align}
Now since $2000=6\times 334 -4$, we have that $a_{2000}=-1$.
A: The first equation is equivalent to $x^2-x+1 = 0$ so $x = (1 \pm \sqrt{-3})/2 = \exp(\pm i \pi/3)$
Taking an exponential to a power is a simple multiplication
$(\exp(\pm i \pi/3)^m = (\exp(\pm i m\pi/3))$
If $n$ is an integer  $\exp(i 2n\pi) = 1$ and $\exp(i (2n-1)\pi) = -1$.
Substituting back in gives $x^{2000} + x^{-2000} = e^{\pm i \pi (2/3 + 666 )} + e^{\mp i \pi(2/3+666)} =  e^{\pm i \pi 2/3} + e^{\mp i \pi 2/3} = 2\cos(2 \pi/3) = -1$
