Let α and β be the roots of $x^2-6x-2=0$.If $a_n={\alpha }^n-{\beta }^{n}forn\ge 1,$then the value of $\frac{a_{10}-2a_8}{3a_9}$ is

 Since, α, β are the roots of the equation
$$\begin{gathered} x^2-6x-2=0\Rightarrow {\alpha }^2-6\alpha -2=0 \\ multiplying both sides by {\alpha }^8 \\ \Rightarrow {\alpha }^{10}-6{\alpha }^9-2{\alpha }^8=0\to \left(1\right) \end{gathered}$$
Similarly, ${\beta }^{10}-6{\beta }^9-2{\beta }^8=0\to \left(2\right)$
Subtracting (2) from (1), we get 
$$\begin{gathered} ({\alpha }^{10}-{\beta }^{10})-6({\alpha }^9-{\beta }^9)-2({\alpha }^8-{\beta }^8)=0 \\ \Rightarrow a_{10}-6a_9-2a_8=0 \\ \Rightarrow \frac{a_{10}-2a_8}{3a_9}=2 \end{gathered}$$