If the sum of the squares of the reciprocals of the roots $\alpha$ and $\beta$ of

the equation 3x² + $\lambda$x $-$ 1 = 0 is 15, then 6($\alpha$³ + $\beta$³)² is equal to :

<p>$3{x^2} + \lambda x - 1 = 0$</p> <p>Given, two roots are $\alpha$ and $\beta$.</p> <p>$\therefore$ Sum of roots $= \alpha + \beta = {-\lambda \over 3}$</p> <p>And product of roots $= \alpha \beta = {-1 \over 3}$</p> <p>Given that,</p> <p>Sum of square of reciprocal of roots $\alpha$ and $\beta$ is 15.</p> <p>$\therefore$ ${1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} = 15$</p> <p>$\Rightarrow {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} = 15$</p> <p>$$ \Rightarrow {{{{(\alpha + \beta )}^2} - 2\alpha \beta } \over {{{(\alpha \beta )}^2}}} = 15$$</p> <p>$$ \Rightarrow {{{{{\lambda ^2}} \over 9} + 2 \times {1 \over 3}} \over {{1 \over 9}}} = 15$$</p> <p>$\Rightarrow {{{{{\lambda ^2} + 6} \over 9}} \over {{1 \over 9}}} = 15$</p> <p>$\Rightarrow {\lambda ^2} + 6 = 15$</p> <p>$\Rightarrow {\lambda ^2} = 9$</p> <p>Now, $6{({\alpha ^3} + {\beta ^3})^2}$</p> <p>$= 6{\{ (\alpha + \beta )({\alpha ^2} + {\beta ^2} - \alpha \beta )\} ^2}$</p> <p>$$ = 6{(\alpha + \beta )^2}{\left[ {{{(\alpha + \beta )}^2} - 2\alpha \beta - \alpha \beta } \right]^2}$$</p> <p>$$ = 6{\left( {{-\lambda \over 3}} \right)^2}{\left[ {{{\left( {{-\lambda \over 3}} \right)}^2} - 3\,.\,{-1 \over 3}} \right]^2}$$</p> <p>$$ = 6 \times {{{\lambda ^2}} \over 9} \times \left[ {{{{\lambda ^2}} \over 9} + 1} \right]$$</p> <p>$= 6 \times {{9} \over 9} \times {\left[ {{{9} \over 9} + 1} \right]^2}$</p> <p>$= 6 \times {\left( {{2}} \right)^2}$</p> <p>$= {{6 \times 4}} = 24$</p>