Let  $\alpha$ and $\beta$ be the roots of equation X² - 6x - 2 = O. If $a_n={\alpha }^n-{\beta }^n,\text{ for }n\ge 1$,then the value $\frac{{\mathrm{a}}_{10}-2{\mathrm{a}}_8}{2{\mathrm{a}}_9}$ as is equal to

Given,$\alpha$ and $\beta$ are the roots of the equation
$x^2-6x-2=0$
$\begin{array}{r}\because {\mathrm{a}}_{\mathrm{n}}={\mathrm{\alpha }}^{\mathrm{n}}-{\mathrm{\beta }}^{\mathrm{n}}\text{ for }\mathrm{n}\ge 1\\ \therefore {\mathrm{a}}_{10}={\mathrm{\alpha }}^{10}-{\mathrm{\beta }}^{10}\\ {\mathrm{a}}_8={\mathrm{\alpha }}^8-{\mathrm{\beta }}^8\Rightarrow {\mathrm{a}}_9={\mathrm{\alpha }}^9-{\mathrm{\beta }}^9\end{array}$
Now, consider
$\frac{{\mathrm{a}}_{10}-2{\mathrm{a}}_8}{2{\mathrm{a}}_9}=\frac{{\mathrm{\alpha }}^{10}-{\mathrm{\beta }}^{10}-2\left({\mathrm{\alpha }}^8-{\mathrm{\beta }}^8\right)}{2\left({\mathrm{\alpha }}^9-{\mathrm{\beta }}^9\right)}$
$\left.\begin{array}{l}=\frac{{\mathrm{\alpha }}^8\left({\mathrm{\alpha }}^2-2\right)-{\mathrm{\beta }}^8\left({\mathrm{\beta }}^2-2\right)}{2\left({\mathrm{\alpha }}^9-{\mathrm{\beta }}^9\right)}\\ =\frac{{\mathrm{\alpha }}^8\cdot 6\mathrm{\alpha }-{\mathrm{\beta }}^8\cdot 6\mathrm{\beta }}{2\left({\mathrm{\alpha }}^9-{\mathrm{\beta }}^9\right)}\\ =\frac{6{\mathrm{\alpha }}^9-6{\mathrm{\beta }}^9}{2\left({\mathrm{\alpha }}^9-{\mathrm{\beta }}^9\right)}\\ =\frac{6}{2}=3\end{array}\mid \begin{array}{c}\because \mathrm{\alpha }\text{ and }\mathrm{\beta }\text{ are the }\\ \text{ roots of }\\ {\mathrm{x}}^2-6\mathrm{x}-2=0\\ \text{ or }{\mathrm{x}}^2=6\mathrm{x}+2\\ \Rightarrow {\mathrm{\alpha }}^2=6\mathrm{\alpha }+2\\ \Rightarrow {\mathrm{\alpha }}^2-2=6\mathrm{\alpha }\\ \text{ and }{\mathrm{\beta }}^2=6\mathrm{\beta }+2\\ \Rightarrow {\mathrm{\beta }}^2-2=6\mathrm{\beta }\end{array}\right]$