Sum of cubes of roots

Given: $$3x^2 + \lambda x - 1 = 0$$ with roots $$\alpha$$ and $$\beta$$.

By Vieta's formulas:

$$ \alpha + \beta = -\frac{\lambda}{3}, \quad \alpha\beta = -\frac{1}{3} $$

We are given that the sum of the squares of the reciprocals of the roots is 15:

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

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

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

And $$\alpha^2\beta^2 = (\alpha\beta)^2 = \frac{1}{9}$$

Substituting:

$$ \frac{\frac{\lambda^2}{9} + \frac{2}{3}}{\frac{1}{9}} = 15 $$

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

$$ \lambda^2 + 6 = 15 $$

$$ \lambda^2 = 9 \implies \lambda = \pm 3 $$

Now we need to find $$6(\alpha^3 + \beta^3)^2$$.

Using the identity: $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$

$$ \alpha + \beta = -\frac{\lambda}{3} = \pm\frac{3}{3} = \pm 1 $$

$$ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) $$

$$ = (\pm 1)^3 - 3\left(-\frac{1}{3}\right)(\pm 1) $$

$$ = \pm 1 + (\pm 1) = \pm 2 $$

Therefore:

$$ 6(\alpha^3 + \beta^3)^2 = 6 \times (\pm 2)^2 = 6 \times 4 = 24 $$

The answer is Option C: 24.