If the area enclosed by the parabolas $$P_1: 2y = 5x^2$$ and $$P_2: x^2 - y + 6 = 0$$ is equal to the area enclosed by $$P_1$$ and $$y = \alpha x$$, $$\alpha > 0$$, then $$\alpha^3$$ is equal to _____.

We are to find $$\alpha^3$$ such that the area enclosed by $$P_1: 2y = 5x^2$$ and $$P_2: x^2 - y + 6 = 0$$ equals the area enclosed by $$P_1$$ and the line $$y = \alpha x$$ with $$\alpha > 0$$.

Since we first rewrite $$P_1$$ and $$P_2$$ as $$y = \frac{5x^2}{2}$$ and $$y = x^2 + 6$$, respectively, their points of intersection satisfy $$\frac{5x^2}{2} = x^2 + 6 \Rightarrow 3x^2 = 12 \Rightarrow x = \pm 2$$. Hence the area between these curves is

$$ A_1 = \int_{-2}^{2}\Bigl(x^2 + 6 - \frac{5x^2}{2}\Bigr)\,dx = \int_{-2}^{2}\Bigl(6 - \frac{3x^2}{2}\Bigr)\,dx = 2\int_0^2\Bigl(6 - \frac{3x^2}{2}\Bigr)\,dx = 2\Bigl[6x - \frac{x^3}{2}\Bigr]_0^2 = 2(12 - 4) = 16. $$

Next, to find the area between $$P_1$$ and the line $$y = \alpha x$$, we solve $$\frac{5x^2}{2} = \alpha x \;\Rightarrow\; x\Bigl(\frac{5x}{2} - \alpha\Bigr)=0 \;\Rightarrow\; x = 0,\;\frac{2\alpha}{5}$$. Therefore,

$$ A_2 = \int_{0}^{2\alpha/5}\Bigl(\alpha x - \frac{5x^2}{2}\Bigr)\,dx = \Bigl[\frac{\alpha x^2}{2} - \frac{5x^3}{6}\Bigr]_0^{2\alpha/5} = \frac{\alpha}{2}\cdot\frac{4\alpha^2}{25} - \frac{5}{6}\cdot\frac{8\alpha^3}{125} = \frac{2\alpha^3}{25} - \frac{4\alpha^3}{75} = \frac{2\alpha^3}{75}. $$

From the above, setting these areas equal gives $$\frac{2\alpha^3}{75} = 16$$, which leads to

$$ \alpha^3 = \frac{16 \times 75}{2} = 600. $$

Therefore, the value of $$\alpha^3$$ is 600.