If the sum of the squares of the reciprocals of the roots α and β...

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

For equation $3{x}^{2} + \lambdax - 1 = 0$

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

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

$⇒ {\alpha}^{2} + {\beta}^{2} = \dfrac{{\lambda}^{2} + 6}{9}$

Now $\dfrac{1}{{\alpha}^{2}} + \dfrac{1}{{\beta}^{2}} = 15 ⇒ \dfrac{{\alpha}^{2} + {\beta}^{2}}{{\alpha}^{2}{\beta}^{2}} = 15$

$⇒ \dfrac{{\lambda}^{2} + 6}{9} \times \dfrac{1}{{\alpha}^{2}{\beta}^{2}} = 15$ $⇒ \dfrac{{\lambda}^{2} + 6}{9} \times \dfrac{1}{\dfrac{1}{9}} = 15$

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

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

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

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