If the sum of the squares of the reciprocals of the roots α and β of the equation 3 x 2 + λ x…

Given $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=15$

For equation $3x^2+\lambda x-1=0$

 $\alpha +\beta =\frac{-\lambda }{3}$, $\alpha \beta =\frac{-1}{3}\Rightarrow {\alpha }^2{\beta }^2=\frac{1}{9}$

then ${\alpha }^2+{\beta }^2={\left(\alpha +\beta \right)}^2-2\alpha \beta$$=\frac{{\lambda }^2}{9}-2\left(\frac{-1}{3}\right)=\frac{{\lambda }^2}{9}+\frac{2}{3}$

 $\Rightarrow {\alpha }^2+{\beta }^2=\frac{{\lambda }^2+6}{9}$

Now $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=15\Rightarrow \frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}=15$

$\Rightarrow \frac{{\lambda }^2+6}{9}\times \frac{1}{{\alpha }^2{\beta }^2}=15$  $\Rightarrow \frac{{\lambda }^2+6}{9}\times \frac{1}{{\displaystyle \frac{1}{9}}}=15$

$\Rightarrow {\lambda }^2+6=15$ $\Rightarrow \lambda =\pm 3$

Now $6{\left({\alpha }^3+{\beta }^3\right)}^2$ $=6{\left(\left(\alpha +\beta \right)\left({\alpha }^2-\alpha \beta +{\beta }^2\right)\right)}^2$

$=6\times 1\times {\left(\frac{15}{9}+\frac{1}{3}\right)}^2$

$=6\times 1\times {\left(2\right)}^2=6\times 4=24$