What will be the solubility of $\mathrm{A}{\left(\mathrm{OH}\right)}_3$ in its aqueous solution if its solubility product is $2.7\times {10}^{-39}?$

$\mathrm{A}\left(\mathrm{OH}{)}_3\right(s)⟶\underset{\mathrm{s}}{A^{3+}\left(aq\right)}+\underset{\mathit{3}\mathit{s}\mathit{ }\mathit{+}\mathit{x}}{\mathit{3}{\mathit{OH}}^{\mathit{-}}\mathit{(}\mathit{a}\mathit{q}})K_{sp}=s(3s+x{)}^3$

$$\begin{gathered} 2{\mathrm{H}}_2\mathrm{O}\left(l\right)⟶\underset{\mathrm{x}}{{\mathrm{H}}_3{\mathrm{O}}^{+}\left(aq\right)}+\underset{x+3s}{{\mathrm{OH}}^{-}\left(aq\right)} \\ {K}_w=(x+3s)\cdot x \end{gathered}$$

$\begin{array}{l}2.7\times {10}^{-39}=s\times {10}^{-21}\\ s=2.7\times {10}^{-18}\mathrm{M}\text{ litre }\\ s=2.7\times {10}^{-15}\mathrm{mole}/{\mathrm{m}}^3\end{array}$