Let $\mathrm{\alpha }$ and $\mathrm{\beta }$ be the roots of ${\mathrm{x}}^2-6\mathrm{x}-2=0$, with $\mathrm{\alpha }>\mathrm{\beta }$. If ${\mathrm{a}}_{\mathrm{n}}={\mathrm{\alpha }}^{\mathrm{n}}-{\mathrm{\beta }}^{\mathrm{n}}\text{ for }\mathrm{n}\ge 1$, then the value of $\frac{{\mathrm{a}}_{10}-2{\mathrm{a}}_8}{2{\mathrm{a}}_9}$ is

${\mathrm{a}}_{\mathrm{n}}={\mathrm{\alpha }}^{\mathrm{n}}-{\mathrm{\beta }}^{\mathrm{n}}$

Also ${\mathrm{\alpha }}^2-6\mathrm{\alpha }-2=0$

Multiply with ${\mathrm{\alpha }}^8$ on both sides

$\Rightarrow {\mathrm{\alpha }}^{10}-6{\mathrm{\alpha }}^9-2{\mathrm{\alpha }}^8=0$                …… (1)

similarly ${\mathrm{\beta }}^{10}-6{\mathrm{\beta }}^9-2{\mathrm{\beta }}^8=0$           ……. (2)

Subtracting (2) from (1) we have

$\begin{array}{l}{\mathrm{\alpha }}^{10}-{\mathrm{\beta }}^{10}-6\left({\mathrm{\alpha }}^9-{\mathrm{\beta }}^9\right)=2\left({\mathrm{\alpha }}^8-{\mathrm{\beta }}^8\right)\\ \Rightarrow {\mathrm{a}}_{10}-6{\mathrm{a}}_9=2{\mathrm{a}}_8\\ \Rightarrow \frac{{\mathrm{a}}_{10}-2{\mathrm{a}}_8}{2{\mathrm{a}}_9}=3.\end{array}$