$\alpha,\beta$

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

Solution without solving for $x$ .

$\begin{aligned} x^2 - \sqrt 3 x + 1 & = 0 \\ x^2 & = \sqrt 3 x - 1 \\ x^3 & = \sqrt 3 x^2 - x \\ & = \sqrt 3\left(\sqrt 3 x - 1\right) - x \\ & = 2x - \sqrt 3 \\ x^5 & = \left(\sqrt 3 x - 1\right)\left(2x - \sqrt 3 x\right) \\ & = 2\sqrt 3 x^2 - 5x + \sqrt 3 \\ & = 2\sqrt 3 \left(\sqrt 3 x - 1\right) - 5x + \sqrt 3 \\ & = x - \sqrt 3 \\ x^{10} & = \left(x - \sqrt 3 \right)^ 2 \\ & = x^2 - 2\sqrt 3 x + 3 \\ & = {\color{#3D99F6} x^2 - \sqrt 3 x + 1} - 2\sqrt 3 + 2 & \small \color{#3D99F6} \text{Note that } x^2 - \sqrt 3 x + 1 = 0 \\ & = 2 - \sqrt 3x \end{aligned}$

Since $\alpha$ and $\beta$ are roots of $x^2 - \sqrt 3 x + 1 = 0$ , then

$\begin{aligned} \alpha^{10} & = 2 - \sqrt 3 \alpha \\ \beta^{10} & = 2 - \sqrt 3 \beta \\ \implies \alpha^{10}+ \beta^{10} & = 4 - \sqrt 3 ({\color{#3D99F6}\alpha + \beta}) & \small \color{#3D99F6} \text{By Vieta's formula} \\ & = 4 - \sqrt 3 ({\color{#3D99F6}\sqrt 3}) \\ & = \boxed{1} \end{aligned}$