Symmetric Expression of Quadratic Roots

For $3x^2 + \lambda x - 1 = 0$: $\alpha + \beta = -\frac{\lambda}{3}, \quad \alpha \beta = -\frac{1}{3}$

Given: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = 15$ $(\alpha+\beta)^2 - 2\left(-\frac{1}{3}\right) = 15 \left(-\frac{1}{3}\right)^2$ $(\alpha+\beta)^2 + \frac{2}{3} = \frac{15}{9} = \frac{5}{3}$ $(\alpha+\beta)^2 = 1 \implies \alpha + \beta = \pm 1$

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

Thus, $(\alpha^3 + \beta^3)^2 = (\pm 2)^2 = 4$.