Let $\alpha$ and $\beta$ be the roots of x² $-$ 6x $-$ 2 = 0. If a_n = $\alpha$ⁿ $-$ $\beta$ⁿ for n $\ge$ 1, then the value of ${{{a_{10}} - 2{a_8}} \over {3{a_9}}}$ is :

Given, $\alpha$ and $\beta$ be the roots of ${x^2} - 6x - 2 = 0$<br><br>$\matrix{ {\alpha + \beta = 6} \cr {\alpha \beta = - 2} \cr }$<br><br>and ${\alpha ^2} - 6\alpha - 2 = 0 \Rightarrow {\alpha ^2} - 2 = 6\alpha$<br><br>${\beta ^2} - 6\beta - 2 = 0 \Rightarrow {\beta ^2} - 2 = 6\beta$<br><br>$${{{a_{10}} - 2{a_8}} \over {3{a_9}}} = {{\left( {{\alpha ^{10}} - {\beta ^{10}}} \right) - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$<br><br>$$ = {{\left( {{\alpha ^{10}} - 2{\alpha ^8}} \right) - \left( {{\beta ^{10}} - 2{\beta ^8}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$<br><br>Now, $$ = {{{\alpha ^8}\left( {{\alpha ^2} - 2} \right) - {\beta ^8}\left( {{\beta ^2} - 2} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$<br><br>$$ = {{{\alpha ^8}(6\alpha ) - {\beta ^8}(6\beta )} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}} = {{6\left( {{\alpha ^9} - {\beta ^9}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}} = {6 \over 3} = 2$$