Let x= sin (2 tan^-1 α) and y= sin (1/2 tan^-1 4/3). If S={α ∈ R: y^2=1-x}, then Σ_α ∈ S 16 α^3 is equal to \ \ \ \.

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