Let $\alpha$ and $\beta$ be the roots of $x^2-6x-2=0$, with $\alpha >\beta$. If $a_n={\alpha }^n-{\beta }^n$ for $n\ge 1$, the value of $\frac{a_{10}-2a_8}{2a_9}$ is 

Given equation is,

$x^2-6x-2=0$

Since,

$$\begin{gathered} {\alpha }^2-6\alpha -2=0 \\ \Rightarrow {\alpha }^2-2=6\alpha .....\left(i\right) \end{gathered}$$

and ${\beta }^2-6\beta -2=0\Rightarrow {\beta }^2-2=6\beta ......\left(ii\right)$

Thus,

$$\begin{gathered} \frac{a_{10}-2a_8}{2a_9}=\frac{({\alpha }^{10}-{\beta }^{10})-2({\alpha }^8-{\beta }^8)}{2({\alpha }^9-{\beta }^9)} \\ =\frac{{\alpha }^8({\alpha }^2-2)-{\beta }^8({\beta }^2-2)}{2({\alpha }^9-{\beta }^9)} \end{gathered}$$

$=\frac{{\alpha }^8. 6\alpha -{\beta }^8. 6\beta }{2({\alpha }^9-{\beta }^9)}$   [From Eqs.(i) and (ii)]

$=\frac{6({\alpha }^9-{\beta }^9)}{2({\alpha }^9-{\beta }^9)}=3$