Let x × y = x ^ 2 + y ^ 3 and (x × 1) × 1 = x ×(1 × 1). Then a...

Given $x \times y = {x}^{2} + {y}^{3}$

Now \left(x \times 1\right) \times 1 = x \times $\left(1 \times 1\right)$

\left({x}^{2} + {1}^{3}\right) \times 1 = x \times $\left({1}^{2} + {1}^{3}\right)$

${\left({x}^{2} + 1\right)}^{2} + {1}^{3} = {x}^{2} + {2}^{3}$

${x}^{4} + 1 + 2{x}^{2} + 1 = {x}^{2} + 8$

${x}^{4} + {x}^{2} - 6 = 0$

Let ${x}^{2} = t$

${t}^{2} + t - 6 = 0$

\left(t + 3\right)\left(t - 2\right) = 0$</p><p>$t \ne - 3$so$t = 2 or ${x}^{2} = 2$

Now $2\sin^{- 1}\left(\dfrac{{x}^{4} + {x}^{2} - 2}{{x}^{4} + {x}^{2} + 2}\right) = 2\sin^{- 1}\left(\dfrac{4 + 2 - 2}{4 + 2 + 2}\right)$
$= 2\sin^{- 1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{6} \times 2 = \dfrac{\pi}{3}$