Explanation

(d)
We have, $(\sin^{- 1}{x\left. \ \ \right)^{3} + (\cos^{- 1}{x\left. \ \ \right)^{3}\ }}$
$= \left( \sin^{- 1}{+ \cos^{- 1}x} \right)^{3} - 3\sin^{- 1}{x\cos^{- 1}{x\ (\sin^{- 1}{x + \cos^{- 1}{x)}}}}$
$= \frac{\pi^{3}}{8} - 3(\sin^{- 1}{x\cos^{- 1}{x)\frac{\pi}{2}}}$
$= \frac{\pi^{3}}{8} - \frac{3\pi}{2}\sin^{- 1}x\left( \frac{\pi}{2} - \sin^{- 1}x \right)$
$= \frac{\pi^{3}}{8} - \frac{3\pi}{2}\sin^{- 1}{x +}\frac{3\pi}{2}(\sin^{- 1}{x\left. \ \ \right)^{2}}$
$= \frac{\pi^{3}}{8} + \frac{3\pi^{\ }}{2}\left\lbrack \left( \sin^{- 1}x \right)^{2} - \frac{\pi}{2}\sin^{- 1}x \right\rbrack$
$= \frac{\pi^{3}}{8} + \frac{3\pi}{2}\left\lbrack \left( \sin^{- 1}x \right)^{2} - \frac{\pi}{2}\sin^{- 1}x + \frac{\pi^{2}}{16} - \frac{\pi^{2}}{16} \right\rbrack$
$= \frac{\pi^{3}}{8} + \frac{3\pi}{2}\left\lbrack \left( \sin^{- 1}{x -}\frac{\pi}{4} \right)^{2} \right\rbrack - \frac{3\pi^{3}}{32}$
$= \frac{\pi^{3}}{32} + \frac{3\pi}{2}\left( \sin^{- 1}{x -}\frac{\pi}{4} \right)^{2}$
$\therefore$ The least value is $\frac{\pi^{3}}{32}$
Since, $\left( \sin^{- 1}{x - \frac{\pi}{4}} \right)^{2} \leq \left( \frac{3\pi}{4} \right)^{2}$
$\therefore$ The greatest value is $\frac{\pi^{3}}{32} + \frac{9\pi^{2}}{16} \times \frac{3\pi}{2} = \frac{7\pi^{3}}{8}$