Let ABC be an isosceles triangle in which A is at $$(-1, 0)$$, $$\angle A = \frac{2\pi}{3}$$, $$AB = AC$$ and B is on the positive x-axis. If $$BC = 4\sqrt{3}$$ and the line BC intersects the line $$y = x + 3$$ at $$(\alpha, \beta)$$, then $$\frac{\beta^4}{\alpha^2}$$ is:

We have an isosceles triangle ABC with $$A(-1, 0)$$, $$\angle A = \frac{2\pi}{3}$$, $$AB = AC$$, and B on the positive x-axis. Also $$BC = 4\sqrt{3}$$.

Since $$\angle A = 120°$$ and $$AB = AC$$, the base angles are each $$\frac{180° - 120°}{2} = 30°$$.

Using the sine rule: $$\frac{BC}{\sin A} = \frac{AB}{\sin C}$$

$$\frac{4\sqrt{3}}{\sin 120°} = \frac{AB}{\sin 30°}$$

$$\frac{4\sqrt{3}}{\frac{\sqrt{3}}{2}} = \frac{AB}{\frac{1}{2}}$$

$$8 = 2 \cdot AB$$

$$AB = 4$$

Since B is on the positive x-axis and $$AB = 4$$, we have $$B = (-1 + 4, 0) = (3, 0)$$.

Now, since the triangle is isosceles with $$AB = AC = 4$$ and $$\angle A = 120°$$, the direction from A to C makes an angle of $$120°$$ with AB.

Direction of AB is along the positive x-axis (angle $$0°$$). So AC is at angle $$120°$$ from AB.

$$C = A + 4(\cos 120°, \sin 120°) = (-1 + 4(-\frac{1}{2}), 0 + 4 \cdot \frac{\sqrt{3}}{2}) = (-3, 2\sqrt{3})$$

Or at angle $$-120°$$: $$C = (-3, -2\sqrt{3})$$.

Let's check $$BC$$: $$BC = \sqrt{(3-(-3))^2 + (0-2\sqrt{3})^2} = \sqrt{36 + 12} = \sqrt{48} = 4\sqrt{3}$$. Correct!

Line BC passes through $$B(3, 0)$$ and $$C(-3, 2\sqrt{3})$$:

Slope = $$\frac{2\sqrt{3} - 0}{-3 - 3} = \frac{2\sqrt{3}}{-6} = -\frac{\sqrt{3}}{3}$$

Equation: $$y = -\frac{\sqrt{3}}{3}(x - 3) = -\frac{\sqrt{3}}{3}x + \sqrt{3}$$

Intersection with $$y = x + 3$$:

$$x + 3 = -\frac{\sqrt{3}}{3}x + \sqrt{3}$$

$$x + \frac{\sqrt{3}}{3}x = \sqrt{3} - 3$$

$$x\left(1 + \frac{\sqrt{3}}{3}\right) = \sqrt{3} - 3$$

$$x \cdot \frac{3 + \sqrt{3}}{3} = \sqrt{3} - 3$$

$$x = \frac{3(\sqrt{3} - 3)}{3 + \sqrt{3}} = \frac{3(\sqrt{3} - 3)(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{3(\sqrt{3} - 3)(3 - \sqrt{3})}{9 - 3} = \frac{3(\sqrt{3} - 3)(3 - \sqrt{3})}{6}$$

$$(\sqrt{3} - 3)(3 - \sqrt{3}) = -(3 - \sqrt{3})(3 - \sqrt{3}) = -(3 - \sqrt{3})^2 = -(9 - 6\sqrt{3} + 3) = -(12 - 6\sqrt{3})$$

$$x = \frac{3 \cdot (-(12 - 6\sqrt{3}))}{6} = \frac{-3(12 - 6\sqrt{3})}{6} = \frac{-36 + 18\sqrt{3}}{6} = -6 + 3\sqrt{3}$$

$$\beta = x + 3 = -6 + 3\sqrt{3} + 3 = -3 + 3\sqrt{3} = 3(\sqrt{3} - 1)$$

$$\alpha = -6 + 3\sqrt{3} = 3(\sqrt{3} - 2)$$

For the other case $$C = (-3, -2\sqrt{3})$$, slope = $$\frac{\sqrt{3}}{3}$$, equation: $$y = \frac{\sqrt{3}}{3}(x-3) = \frac{\sqrt{3}}{3}x - \sqrt{3}$$.

Intersection: $$x + 3 = \frac{\sqrt{3}}{3}x - \sqrt{3}$$, so $$x(1 - \frac{\sqrt{3}}{3}) = -\sqrt{3} - 3$$.

$$x = \frac{-3(\sqrt{3}+3)}{3-\sqrt{3}} \cdot \frac{3+\sqrt{3}}{3+\sqrt{3}} = \frac{-3(\sqrt{3}+3)(3+\sqrt{3})}{6}$$

$$= \frac{-3(3\sqrt{3}+3+9+3\sqrt{3})}{6} = \frac{-3(12+6\sqrt{3})}{6} = -(6+3\sqrt{3})$$

$$\alpha = -6-3\sqrt{3}$$, $$\beta = -3-3\sqrt{3}$$.

Now compute $$\frac{\beta^4}{\alpha^2}$$. Using the first case:

$$\alpha = 3(\sqrt{3}-2)$$, $$\beta = 3(\sqrt{3}-1)$$

$$\frac{\beta^4}{\alpha^2} = \frac{81(\sqrt{3}-1)^4}{9(\sqrt{3}-2)^2} = \frac{9(\sqrt{3}-1)^4}{(\sqrt{3}-2)^2}$$

$$(\sqrt{3}-1)^2 = 4-2\sqrt{3}$$, so $$(\sqrt{3}-1)^4 = (4-2\sqrt{3})^2 = 16-16\sqrt{3}+12 = 28-16\sqrt{3}$$

$$(\sqrt{3}-2)^2 = 7-4\sqrt{3}$$

$$\frac{9(28-16\sqrt{3})}{7-4\sqrt{3}} = \frac{9 \cdot 4(7-4\sqrt{3})}{7-4\sqrt{3}} = 36$$

The answer is $$\boxed{36}$$.