Consider two circles $$C_1: x^2 + y^2 = 25$$ and $$C_2: (x-\alpha)^2 + y^2 = 16$$, where $$\alpha \in (5, 9)$$. Let the angle between the two radii (one to each circle) drawn from one of the intersection points of $$C_1$$ and $$C_2$$ be $$\sin^{-1}\frac{\sqrt{63}}{8}$$. If the length of common chord of $$C_1$$ and $$C_2$$ is $$\beta$$, then the value of $$(\alpha\beta)^2$$ equals _________.

$$C_1: x^2+y^2=25$$ (center O₁(0,0), r₁=5). $$C_2: (x-\alpha)^2+y^2=16$$ (center O₂(α,0), r₂=4).

The angle between radii at intersection point is $$\sin^{-1}\frac{\sqrt{63}}{8}$$.

Let the angle be $$\theta$$ at the intersection point P. Using the law of cosines in triangle O₁PO₂:

$$\alpha^2 = 25 + 16 - 2(5)(4)\cos\theta = 41 - 40\cos\theta$$

where $$\theta$$ is the angle at P. $$\sin\theta = \frac{\sqrt{63}}{8}$$, $$\cos\theta = \pm\frac{1}{8}$$.

Since $$\alpha \in (5,9)$$, $$\alpha^2 \in (25,81)$$. With $$\cos\theta = 1/8$$: $$\alpha^2 = 41-5 = 36$$, $$\alpha = 6$$ ✓.

Common chord length: The perpendicular from O₁ to the radical axis. Radical axis: subtract circles: $$-2\alpha x + \alpha^2 = -9 \Rightarrow x = \frac{\alpha^2+9}{2\alpha} = \frac{45}{12} = \frac{15}{4}$$.

$$\beta = 2\sqrt{r_1^2 - x^2} = 2\sqrt{25 - 225/16} = 2\sqrt{175/16} = 2 \cdot \frac{5\sqrt{7}}{4} = \frac{5\sqrt{7}}{2}$$.

$$(\alpha\beta)^2 = (6 \cdot \frac{5\sqrt{7}}{2})^2 = (15\sqrt{7})^2 = 225 \times 7 = 1575$$.

Therefore, the answer is $$\boxed{1575}$$.