If $\alpha \&\beta$ be the roots of $x^2-6x-2=0$ with $\alpha >\beta .$ If $a_n={\alpha }^n-{\beta }^n, (n>1),$ then the value of $\frac{a_{10}-2a_8}{2a_9}-2=$ ______.

$x^2-6x-2=0$ has roots $(a,\beta )$  
${\alpha }^2-6\alpha -2=0$ and ${\beta }^2-6\beta -2=0$  
$\frac{a_{10}-2a_8}{2a_9}=\frac{({\alpha }^{10}-{\beta }^{10})-2({\alpha }^8-{\beta }^8)}{2({\alpha }^9-{\beta }^9)}=\frac{{\alpha }^8({\alpha }^2-2)-{\beta }^8({\beta }^2-2)}{2({\alpha }^9-{\beta }^9)}$ 
$=\frac{{\alpha }^8\left(6\alpha \right)-{\beta }^8\left(6\beta \right)}{2({\alpha }^9-{\beta }^9)}=\frac{6({\alpha }^9-{\beta }^9)}{2({\alpha }^9-{\beta }^9)}=3$