If the sum of the squares of the reciprocals of the roots α and ß of the equation $3{\mathrm{x}}^2+\mathrm{\lambda x}-1=0$ is 15, then $6{\left({\mathrm{\alpha }}^3+{\mathrm{\beta }}^3\right)}^2$ is equal to

Here α,ß roots of equation $3{\mathrm{x}}^2+\mathrm{\lambda x}-1=0$
$\begin{array}{l}\mathrm{\alpha }+\mathrm{\beta }=\frac{-\mathrm{\lambda }}{3},\mathrm{\alpha \beta }=\frac{-1}{3}\\ \frac{1}{{\mathrm{\alpha }}^2}+\frac{1}{{\mathrm{\beta }}^2}=\frac{(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }}{{\mathrm{\alpha }}^2{\mathrm{\beta }}^2}=15\\ \frac{\left({\mathrm{\lambda }}^2/9\right)+2/3}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$9$}\right.}}=15\Rightarrow {\mathrm{\lambda }}^2+6=15\\ {\mathrm{\lambda }}^2=9\end{array}$
Now $6{\left({\mathrm{\alpha }}^3+{\mathrm{\beta }}^3\right)}^2=6{\left((\mathrm{\alpha }+\mathrm{\beta })\left((\mathrm{\alpha }+\mathrm{\beta }{)}^2-3\mathrm{\alpha \beta }\right)\right)}^2$
$=6\left(\frac{{\mathrm{\lambda }}^2}{9}\right){\left\{\frac{{\mathrm{\lambda }}^2}{9}+1\right\}}^2=24$