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