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

Given  that $\alpha ,\beta$  are the roots of the equation  $x^2-6x-2=0$

Hence,

${\alpha }^2-6\alpha -2=0,{\beta }^2-6\beta -2=0$

${\alpha }^{10}=6{\alpha }^9+2{\alpha }^8,{\beta }^{10}=6{\beta }^9+2{\beta }^8$     (  multiply first equation by ${\alpha }^8$,and  second equation by ${\beta }^8$ on both sides)

 $\begin{array}{c}{a}_{10}={\alpha }^{10}-{\beta }^{10}=6({\alpha }^9-{\beta }^9)+2({\alpha }^8-{\beta }^8)\\ =6a_9+2a_8\\ a_{10}-2a_8=6a_9\\ \frac{a_{10}-2a_8}{3a_9}=2\end{array}$