Let $x = \sin (2{\tan ^{ - 1}}\alpha )$ and $y = \sin \left( {{1 \over 2}{{\tan }^{ - 1}}{4 \over 3}} \right)$. If $S = \{ a \in R:{y^2} = 1 - x\}$, then $\sum\limits_{\alpha \in S}^{} {16{\alpha ^3}}$ is equal to _______________.

<p>$\because$ $$x = \sin \left( {2{{\tan }^{ - 1}}\alpha } \right) = {{2\alpha } \over {1 + {\alpha ^2}}}$$ ...... (i)</p> <p>and $$y = \sin \left( {{1 \over 2}{{\tan }^{ - 1}}{4 \over 3}} \right) = \sin \left( {{{\sin }^{ - 1}}{1 \over {\sqrt 5 }}} \right) = {1 \over {\sqrt 5 }}$$</p> <p>Now, ${y^2} = 1 - x$</p> <p>${1 \over 5} = 1 - {{2\alpha } \over {1 + {\alpha ^2}}}$</p> <p>$\Rightarrow 1 + {\alpha ^2} = 5 + 5{\alpha ^2} - 10\alpha$</p> <p>$\Rightarrow 2{\alpha ^2} - 5\alpha + 2 = 0$</p> <p>$\therefore$ $\alpha = 2,{1 \over 2}$</p> <p>$\therefore$ $$\sum\limits_{\alpha \in S} {16{\alpha ^3} = 16 \times {2^3} + 16 \times {1 \over {{2^3}}} = 130} $$</p>