If the sum of the squares of the reciprocals of the roots \( \alpha \) and \( \beta \) of the equation \( 3x^{2} + \lambda x - 1 = 0 \) is \( 15 \), then \( 6(\alpha^{3} + \beta^{3})^{2} \) is equal to:
\( 18 \)
\( 24 \)
\( 36 \)
\( 96 \)