$\alpha ,\beta$ are roots of $x^2-6x-2=0$ with $\alpha >\beta$ and $a_n={\alpha }^n-{\beta }^n$ for $n\ge 1$ then $\frac{a_{10}-2a_8}{2a_9}=$

$\alpha$ is a root of $x^2-6x-2=0$ 
$\Rightarrow {\alpha }^2-6\alpha -2=0$ 
$\begin{array}{l}\Rightarrow {\alpha }^{n+2}-6{\alpha }^{n+1}-2{\alpha }_n=0\to \left(1\right)\\ \Rightarrow {\beta }^{n+2}-6{\beta }^{n+1}-2{\beta }_n=0\to \left(2\right)\end{array}$ 
$\left(1\right)-\left(2\right)\Rightarrow a_n{{ }_{+}}_2-6a_{n+1}-2a_n=0$ 
Put n = 8