Large numbers don't faze me!

Relevant wiki: Vieta's Formula - Forming Quadratics

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

Since $\alpha$ and $\beta$ are roots of $x^2+x+1 =0$ , $\implies \alpha^3 = \beta^3 = 1$ . Therefore,

$\begin{aligned} \alpha^{1027} + \beta^{1027} & = \alpha^{342\times 3 + 1} + \beta^{342\times 3 + 1} \\ & = \alpha + \beta & \small \color{#3D99F6} \text{By Vieta's formula} \\ & = \boxed{-1} \end{aligned}$

Because of Vieta- Cardano formula $\alpha + \beta = - 1, \space \alpha \cdot \beta = 1 \Rightarrow \alpha = \frac{1}{\beta} \Rightarrow$ $\alpha^{1027} + \beta^{1027} = \frac{1}{\beta^{1027}} + \beta^{1027} = \frac{ 1 + \beta^{2054}}{\beta^{1027}}, \space (1)$ Due to $\beta^{1025}\cdot (\beta^2 + \beta + 1) = 0 \Rightarrow \beta^{1027} = - \beta^{1026} - \beta^{1025} = \beta^{1024}$ , keeping on with this process we get $\beta^{1027} = \beta$ and $\beta^{2054} = \beta^2$ . Therefore, $(1) = \frac{ 1 + \beta^{2054}}{\beta^{1027}} = \frac{1 + \beta^2}{\beta} = \frac{-\beta}{\beta} = -1$

Note.- $\beta \neq 0$

Note first that $x^{2} + x + 1 = \dfrac{x^{3} - 1}{x - 1}$ for $x \ne 1$ . Then since $1$ is not a solution to $x^{2} + x + 1 = 0$ we know that the roots $\alpha, \beta$ of this quadratic are such that $\alpha^{3} = 1 = \beta^{3}$ .

Then since $1027 = 3*342 + 1$ we have that $\alpha^{1027} = \alpha$ and $\beta^{1027} = \beta$ , and so by Vieta's

$\alpha^{1027} + \beta^{1027} = \alpha + \beta = \boxed{-1}$ .

Roots are omega and omega square

Alpha = w and beta = w^2 ... Cube roots of unity

a = (-1 + i sqrt(3))/2 = exp(i 2/3 pi)

b = (-1 - i sqrt(3))/2 = exp(-i 2/3 pi)

2q = a^1027 + b^1027

q = cos(1027*2/3 pi) = -0,5

solution = 2q = -1