Let $$x = \sin(2\tan^{-1}\alpha)$$ and $$y = \sin\left(\dfrac{1}{2}\tan^{-1}\dfrac{4}{3}\right)$$. If $$S = \{\alpha \in \mathbb{R} : y^2 = 1 - x\}$$, then $$\displaystyle\sum_{\alpha \in S} 16\alpha^3$$ is equal to ______.

We have $$x = \sin(2\tan^{-1}\alpha)$$ and $$y = \sin\left(\dfrac{1}{2}\tan^{-1}\dfrac{4}{3}\right)$$.

Using the identity $$\sin(2\tan^{-1}\alpha) = \dfrac{2\alpha}{1+\alpha^2}$$, it follows that

$$x = \frac{2\alpha}{1+\alpha^2}$$

Let $$\theta = \tan^{-1}\frac{4}{3}$$, so $$\tan\theta = \frac{4}{3}$$ and $$\cos\theta = \frac{3}{5}$$.

$$y = \sin\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{2}} = \sqrt{\frac{1-\frac{3}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$$

$$y^2 = \frac{1}{5}$$

From $$y^2 = 1 - x$$ and the expressions above, we get

$$\frac{1}{5} = 1 - \frac{2\alpha}{1+\alpha^2}$$

$$\frac{2\alpha}{1+\alpha^2} = \frac{4}{5}$$

$$10\alpha = 4(1+\alpha^2) \implies 4\alpha^2 - 10\alpha + 4 = 0 \implies 2\alpha^2 - 5\alpha + 2 = 0$$

$$\alpha = \frac{5 \pm 3}{4} \implies \alpha = 2 \text{ or } \alpha = \frac{1}{2}$$

The required sum is

$$\sum_{\alpha \in S} 16\alpha^3 = 16\left(2^3 + \left(\frac{1}{2}\right)^3\right) = 16\left(8 + \frac{1}{8}\right) = 16 \times \frac{65}{8} = 130$$

The answer is $$\boxed{130}$$.