It's all about the roots of an equation

$\begin{aligned} \frac {a_{10}-2a_8}{2a_9} & = \frac {\alpha^{10}-\beta^{10}-2(\alpha^8-\beta^8)}{2(\alpha^9-\beta^9)} \\ & = \frac {\alpha^8(\alpha^2-2) - \beta^8(\beta^2-2)}{2(\alpha^9-\beta^9)} & \small \blue{\text{Since }\alpha^2-6\alpha -2 =0 \implies \alpha^2 - 2 = 6\alpha} \\ & = \frac {\alpha^8(6\alpha) - \beta^8(6\beta)}{2(\alpha^9-\beta^9)} & \small \blue{\text{and similarly, } \implies \beta^2 - 2 = 6\beta} \\ & = \frac {6(\alpha^9- \beta^9)}{2(\alpha^9-\beta^9)} \\ & = \boxed 3 \end{aligned}$

We notice that $x^2 - 6x - 2 = 0$ is the characteristic equation of the difference equation $x_{n+2} = 6x_{n+1} + 2x_{n}$ . Since $\alpha$ and $\beta$ are roots of the characteristic equation, each $c_1 \alpha^n + c_2 \beta^n$ is a solution to the difference equation, especially when $c_1 = 1$ and $c_2 = -1$ .

Thus for all $n$ , $\alpha^{n+2} - \beta^{n+2} = 6\left(\alpha^{n+1} - \beta^{n+1}\right) + 2\left(\alpha^{n} - \beta^{n}\right).$ After some easy algebraic manipulation we see that $\dfrac{\alpha^{n+2} - \beta^{n+2} - 2\left(\alpha^{n} - \beta^{n}\right)}{2\left(\alpha^{n+1} - \beta^{n+1}\right)} = \boxed{3}.$