Let the set of all positive values of $$\lambda$$, for which the point of local minimum of the function $$(1 + x(\lambda^2 - x^2))$$ satisfies $$\frac{x^2 + x + 2}{x^2 + 5x + 6} < 0$$, be $$(\alpha, \beta)$$. Then $$\alpha^2 + \beta^2$$ is equal to _________

Find $$\alpha^2 + \beta^2$$ where $$(\alpha, \beta)$$ is the set of positive $$\lambda$$ values for which the local minimum of $$(1 + x(\lambda^2 - x^2))$$ satisfies $$\frac{x^2+x+2}{x^2+5x+6} < 0$$.

$$f'(x) = \lambda^2 - 3x^2 = 0 \Rightarrow x = \pm \frac{\lambda}{\sqrt{3}}$$

$$f''(x) = -6x$$. At $$x = \lambda/\sqrt{3}$$: $$f'' = -6\lambda/\sqrt{3} < 0$$ (local max). At $$x = -\lambda/\sqrt{3}$$: $$f'' = 6\lambda/\sqrt{3} > 0$$ (local min).

The local minimum is at $$x_0 = -\frac{\lambda}{\sqrt{3}}$$ (negative value since $$\lambda > 0$$).

$$\frac{x_0^2 + x_0 + 2}{x_0^2 + 5x_0 + 6} < 0$$

Numerator: $$x_0^2 + x_0 + 2$$. Discriminant = $$1 - 8 = -7 < 0$$. Since leading coefficient > 0, numerator is always positive.

Denominator: $$x_0^2 + 5x_0 + 6 = (x_0+2)(x_0+3) < 0$$

This requires $$-3 < x_0 < -2$$.

$$-3 < -\frac{\lambda}{\sqrt{3}} < -2$$

$$2 < \frac{\lambda}{\sqrt{3}} < 3$$

$$2\sqrt{3} < \lambda < 3\sqrt{3}$$

So $$\alpha = 2\sqrt{3}$$, $$\beta = 3\sqrt{3}$$.

$$\alpha^2 + \beta^2 = 12 + 27 = 39$$

The correct answer is $$\boxed{39}$$.