From the given equation, we have: $\alpha + \beta = -\frac{\lambda}{3}$ and $\alpha\beta = -\frac{1}{3}$.
Using the condition $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15$, we can write this as $\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = 15$.
Substituting $\alpha\beta = -\frac{1}{3}$, we find $\alpha^2 + \beta^2 = -5$.
We use the identity $(\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta)$.
Substituting the known values of $\alpha + \beta$ and $\alpha\beta$, we get $(-\frac{\lambda}{3})^3 = \alpha^3 + \beta^3 + 3(-\frac{1}{3})(-\frac{\lambda}{3})$.
This simplifies to $\alpha^3 + \beta^3 = -\frac{\lambda^3}{27} + \frac{\lambda}{9}$.
To find λ, we use the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$.
Substituting the values, we get $(-\frac{\lambda}{3})^2 = -5 + 2(-\frac{1}{3})$.
Solving for λ, we find $\lambda = \pm 3\sqrt{2}$.
Substituting the value of λ into the expression for $\alpha^3 + \beta^3$, we get $(\alpha^3 + \beta^3)^2 = 4$.
Thus, the value of $(\alpha^3 + \beta^3)^2$ is 4.