Value of Polinomials (3)

Similar solution as @Alak Bhattacharya 's

Note that $x^2 - x + 1 = 0$ has complex roots.

$\begin{aligned} x^2 - x + 1 & = 0 & \small \color{#3D99F6} \text{Rearrange} \\ x^2 & = x - 1 & \small \color{#3D99F6} \text{Multiply both sides by }x \\ x^3 & = {\color{#3D99F6} x^2} - x & \small \color{#3D99F6} \text{Note that }x^2 = x-1 \\ & = {\color{#3D99F6} x-1} - x \\ \implies x^3 & = -1 & \small \color{#3D99F6} \text{Complex root of }-1 \end{aligned}$

Also

$\begin{aligned} x^2 - x + 1 & = 0 & \small \color{#3D99F6} \text{Rearrange} \\ x^2 + 1 & = x & \small \color{#3D99F6} \text{Divide both sides by }x \\ \implies x + \frac 1x & = 1 \end{aligned}$

Therefore,

$\begin{aligned} X & = x^7 + \frac 1{x^7} - 1 \\ & = x(x^3)^2 + \frac 1{x(x^3)^2} - 1 \\ & = x(-1)^2 + \frac 1{x(-1)^2} - 1 \\ & = {\color{#3D99F6} x + \frac 1x} - 1 & \small \color{#3D99F6} \text{Note that }x + \frac 1x = 1 \\ & = {\color{#3D99F6}1} - 1 \\ & = \boxed 0 \end{aligned}$

x is the complex cube root of -1. Therefore the required sum is the sum of all the cube roots of -1 which is zero.