sum of roots

Given that $x^3 - x + 1 = 0$ , $\implies x^3 = x - 1$ . Then

$\begin{aligned} x^{16} & = x(x^3)^5 = x (x-1)^5 \\ & = x(x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1) \\ & = (x-1)^2 - 5x^2(x-1) + 10x(x-1) - 10(x-1) + 5x^2 - x \\ & = x^2 - 2x + 1 - 5(x-1) + 5x^2 + 10x^2 - 10x - 10x + 10 + 5x^2 - x \\ & = 21x^2 - 28x + 16 \end{aligned}$

Since $\alpha$ , $\beta$ , and $\gamma$ are roots of $x^3 - x + 1 =0$ , then

$\begin{aligned} \alpha^{16} + \beta^{16} + \gamma^{16} & = 21 (\alpha^2 + \beta^2 + \gamma^2) - 28 (\alpha + \beta + \gamma) + 16 \times 3 \\ & = 21 \left((\alpha + \beta + \gamma)^2 - 2(\alpha \beta +\beta \gamma + \gamma \alpha \right) - 28(\alpha + \beta + \gamma) + 48 & \small \blue{\text{By Vieta's formula }\alpha + \beta + \gamma = 0} \\ & = 21 \left(0 + 2 \right) - 28(0) + 48 & \small \blue{\text{and }\alpha \beta +\beta \gamma + \gamma \alpha = -1} \\ & = \boxed {90} \end{aligned}$