Very Long Summation

The roots of $x + \dfrac 1x = - 1$ are the cubit roots of unit $\omega$ (see note). Then,

$\begin{aligned} S & = \sum_{n=1}^{333} \left(x^n + \frac 1{x^n} \right) & \small \color{#3D99F6} \text{Since }\omega \text{ is the root.} \\ & = \sum_{n=1}^{333} \left(\omega^n + \omega^{-n} \right) & \small \color{#3D99F6} \text{Two geometric progressions} \\ & = \omega \left(\frac {\omega^{333}-1}{\omega -1}\right) + \omega^{-1} \left(\frac {\omega^{-333}-1}{\omega^{-1} -1}\right) & \small \color{#3D99F6} \text{Since } \omega^3 = 1 \\ & = \boxed 0 \end{aligned}$

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Note:

$\begin{aligned} x + \frac 1x & = - 1 \\ \implies x + 1 + \frac 1x & = 0 & \small \color{#3D99F6} \text{Multiply both sides by }x \\ x^2 + x + 1 & = 0 & \small \color{#3D99F6} \text{Multiply both sides by }x \text{ again.} \\ x^3 + x^2 + x & = 0 & \small \color{#3D99F6} \text{Add 1 to both sides} \\ x^3 + \color{#3D99F6} x^2 + x + 1 & = 1 & \small \color{#3D99F6} \text{Note that } x^2 + x + 1 = 0 \\ \implies x^3 & = 1 \end{aligned}$

If

$x+ \frac{1}{x} = -1$

$x=e^{\frac{2\pi}{3}}$

It can be proved that

$x^n+ \frac{1}{x^{n}} = -1$ ; if n is not a multiple of 3

$x^n+ \frac{1}{x^{n}} = 2$ ; if n is a multiple of 3

so

$\sum\limits_{n=1}^{333} x^n+ \frac{1}{x^{n}} = \frac{333}{3}(-1-1+2) = 0$