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

Given, $\alpha$ is a root of $x^2-6x-2=0$ 
$\therefore {\alpha }^2-6\alpha -2=0$ 
Multiplying by ${\alpha }^n$ it becomes, ${\alpha }^{n+2}-6{\alpha }^{n+1}-2{\alpha }^n=0$  
Similarly, ${\beta }^{n+2}-6{\beta }^{n+1}-2{\beta }^n=0$  
Subtracting, we get $({\alpha }^{n+2}-{\beta }^{n+2})-6({\alpha }^{n+1}-{\beta }^{n+1})-2({\alpha }^n-{\beta }^n)=0$
 
i.e., $a_{n+2}-6a_{n+1}-2a_n=0.$ Thus, $\frac{a_{n+2}-2a_n}{2a_{n+1}}=3$  
set $n=8$ to obtain the desired value $\frac{a_{10}-2a_8}{2a_9}=3$