Let  $\alpha ,\beta$  be roots of $x^2-6x-2=0$   with $\alpha >\beta$ .  If $a_n={\alpha }^n-{\beta }^n$  for $n\ge 1$   then the value of $\frac{a_{10}-2a_8}{3a_9}=\_\_\_\_\_$  

Given  $x^2-6x-2=0$
Roots are  $\alpha ,\beta$
$\Rightarrow \alpha +\beta =6$  and  $\alpha \beta =-2$
If  $\beta =\frac{-2}{\alpha }\Rightarrow \alpha -\frac{2}{\alpha }=6$
$\Rightarrow {\alpha }^2-2=6\alpha$  similarly  ${\beta }^2-2=6\beta$
$a_n={\alpha }^n-{\beta }^n,n\ge 1$ ,  $\frac{a_{10}-2a_8}{3a_9}=\frac{({\alpha }^{10}-{\beta }^{10})-2({\alpha }^8-{\beta }^8)}{3a_9}$

$=\frac{{\alpha }^8(\alpha -2)-{\beta }^8({\beta }^2-2)}{3({\alpha }^9-{\beta }^9)}$   $=\frac{{\alpha }^8\left(6\alpha \right)-{\beta }^8\left(6\beta \right)}{3({\alpha }^9-{\beta }^9)}=\frac{6({\alpha }^9-{\beta }^9)}{3({\alpha }^9-{\beta }^9)}=2$

$\therefore \frac{a_{10}-2a_8}{3a_9}=2$