Let x = sin (2 ^ - 1 α) and y = ((1)/(2) ^ - 1 (4)/(3)) . If S = \...

Given $x = \sin \left(2\tan^{- 1}\alpha\right) = \sin \left(\tan^{- 1}\dfrac{2\alpha}{1 - {\alpha}^{2}}\right) = \sin \left(\sin^{- 1}\dfrac{2\alpha}{1 + {\alpha}^{2}}\right) = \dfrac{2\alpha}{1 + {\alpha}^{2}} ⋯ \left(i\right)$

and $y = \sin \left(\dfrac{1}{2}\tan^{- 1}\dfrac{4}{3}\right) = \sin \left(\tan^{- 1}\dfrac{1}{2}\right) = \sin \left(\sin^{- 1}\dfrac{1}{\sqrt{5}}\right) = \dfrac{1}{\sqrt{5}} . . . \left(ii\right)$

Now, ${y}^{2} = 1 - x$

$⇒ \dfrac{1}{5} = 1 - \dfrac{2\alpha}{1 + {\alpha}^{2}}$ (from \left(i\right) \text{ and } \left(ii\right)$)</p><p>$⇒ 1 + {\alpha}^{2} = 5 + 5{\alpha}^{2} - 10\alpha$</p><p>$⇒ 2{\alpha}^{2} - 5\alpha + 2 = 0$</p><p>$∴ \alpha = 2, \dfrac{1}{2}$</p><p>$∴ \mathop{∑}_{\alpha \in S}16{\alpha}^{3} = 16 \times {2}^{3} + 16 \times \dfrac{1}{{2}^{3}} = 16\left(\dfrac{65}{8}\right)$</p><p>$= 130