If $a<\frac{1}{32},$ then the number of solutions of ${\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3=a{\pi }^3,$ is 

We find that

$\begin{array}{l}{\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3\\ ={\left({\sin }^{-1}x+{\cos }^{-1}x\right)}^3-3{\sin }^{-1}x{\cos }^{-1}x\left({\sin }^{-1}x+{\cos }^{-1}x\right)\end{array}$

$\begin{array}{l}=\frac{{\pi }^3}{8}-3\left({\sin }^{-1}x{\cos }^{-1}x\right)\frac{\pi }{2}\\ =\frac{{\pi }^3}{8}-\frac{3\pi }{2}\left(\frac{\pi }{2}-{\sin }^{-1}x\right){\sin }^{-1}x\\ =\frac{{\pi }^3}{8}-\frac{3{\pi }^2}{4}{\sin }^{-1}x+\frac{3\pi }{2}{\left({\sin }^{-1}x\right)}^2\end{array}$

$\begin{array}{l}=\frac{{\pi }^3}{8}+\frac{3\pi }{2}\left\{{\left({\sin }^{-1}x\right)}^2-\frac{\pi }{2}{\sin }^{-1}x\right\}\\ =\frac{{\pi }^3}{8}+\frac{3\pi }{2}\left\{{\left({\sin }^{-1}x-\frac{\pi }{4}\right)}^2\right\}-\frac{3{\pi }^3}{32}\\ =\frac{{\pi }^3}{32}+\frac{3\pi }{2}{\left({\sin }^{-1}x-\frac{\pi }{4}\right)}^2\end{array}$

$\begin{array}{l}\Rightarrow {\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3\ge \frac{{\pi }^3}{32}\\ \Rightarrow a{\pi }^3\ge \frac{{\pi }^3}{32}\Rightarrow a\ge \frac{1}{32}\end{array}$

But, it is given that $a<\frac{1}{32}.$

Hence, for $a<\frac{1}{32},$ the equation has no solution.