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