$\begin{aligned} & \text { We have }\left(\sin ^{-1} x\right)^3+\left(\cos ^{-1} x\right)^3 \\ & =\left(\sin ^{-1} x+\cos ^{-1} x\right)^3-3 \sin ^{-1} x \cos ^{-1} x\left(\sin ^{-1} x+\cos ^{-1} x\right) \\ & =\frac{\pi^3}{8}-3\left(\sin ^{-1} x \cos ^{-1} x\right) \frac{\pi}{2} \\ & =\frac{\pi^3}{8}-\frac{3 \pi}{2} \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right) \\ & =\frac{\pi^3}{8}-\frac{3 \pi^2}{4} \sin ^{-1} x+\frac{3 \pi}{2}\left(\sin ^{-1} x\right)^2 \\ & =\frac{\pi^3}{8}+\frac{3 \pi}{2}\left[\left(\sin ^{-1} x\right)^2-\frac{\pi}{2} \sin ^{-1} x\right] \\ & =\frac{\pi^3}{8}+\frac{3 \pi}{2}\left[\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2\right]-\frac{3 \pi^3}{32} \\ & =\frac{\pi^3}{32}+\frac{3 \pi}{2}\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2 \\ & \therefore \text { The least value is } \frac{\pi^3}{32} \\ & \text { and since }\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2 \leq\left(\frac{3 \pi}{4}\right)^2 \\ & \therefore \text { The greatest value is } \frac{\pi^3}{32}+\frac{9 \pi^2}{16} \times \frac{3 \pi}{2}=\frac{7 \pi^3}{8} \\ & \end{aligned}$