Find Greatest And Least Value of ${\left({\sin }^{-1} \mathrm{x}\right)}^3+{\left({\cos }^{-1}{x}^{ }\right)}^3$

$$\begin{gathered} Now {\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3={({\sin }^{-1}x}^{ }+{{\cos }^{-1}x)}^3-3{({\sin }^{-1}x}^{ }{\cos }^{-1}x\left)\right({({\sin }^{-1}x}^{ }+{\cos }^{-1}x) \\ ={\left(\frac{\pi }{2}\right)}^3-3\frac{\pi }{2}{\sin }^{-1}x(\frac{\pi }{2}-{\sin }^{-1}x) \\ = \frac{{\pi }^3}{8} -\frac{3{\pi }^2}{4} {\sin }^{-1}x +\frac{3\pi }{2}{( {\sin }^{-1}x )}^2 put \theta ={\sin }^{-1}x and let y=\frac{{\pi }^3}{8}-\frac{3{\pi }^2}{4} \theta +\frac{3\pi }{2}{\theta }^2 \\ \Rightarrow \frac{{\pi }^3}{8}-\frac{3{\pi }^2}{4} \theta +\frac{3\pi }{2}{\theta }^2-y=0 \\ \Rightarrow \frac{3\pi }{2}({\theta }^2-\frac{\pi }{2}\theta +\frac{2}{3\pi }(\frac{{\pi }^3}{8}-y\left)\right)=0 \\ \Rightarrow {\theta }^2-\frac{\theta \pi }{2}+\frac{{\pi }^2}{12}-\frac{2y}{3\pi }=0 \\ \Rightarrow (\theta -\frac{\pi }{4}{)}^2-\frac{{\pi }^2}{16}+\frac{{\pi }^2}{12}-\frac{2y}{3\pi }=0 \\ \Rightarrow {(\theta -\frac{\pi }{4})}^2+\frac{{\pi }^2}{48}-\frac{2y}{3\pi }=0\to (1) \\ We know that \frac{-\pi }{2}\le {\sin }^{-1}{x}^{ }\le \frac{\pi }{2} \\ \Rightarrow \frac{-3\pi }{4}\le {\sin }^{-1}x-\frac{\pi }{4}\le \frac{\pi }{4} \\ \Rightarrow 0\le {({\sin }^{-1}x-\frac{\pi }{4})}^2\le \frac{9{\pi }^2}{16} from (1) \Rightarrow 0\le \frac{2y}{3\pi }-\frac{{\pi }^2}{48}\le \frac{9{\pi }^2}{16} \\ \Rightarrow \frac{{\pi }^2}{48}\le \frac{2y}{3\pi }\le \frac{9{\pi }^2}{16}+\frac{{\pi }^2}{48} \\ \Rightarrow \frac{3\pi }{2}(\frac{{\pi }^2}{48})\le y\le \frac{28{\pi }^2}{48}(\frac{3\pi }{2}) \\ \Rightarrow \frac{{\pi }^3}{32}\le y\le \frac{7\pi }{8} \\ \therefore greatest and least values are \frac{7\pi }{8},\frac{{\pi }^3}{32} \end{gathered}$$