The molar solubility of M(OH)₃ in 0.4 M MX₃ solution in terms of solubility product of M(OH)₃ is

Following equilibria gets simultaneous established

$\mathrm{M}\left(\mathrm{OH}{)}_3\right(\mathrm{s})\rightleftharpoons \mathrm{M}(\underset{\mathrm{S}}{\mathrm{OH}}{)}_3\left(\mathrm{aq}\right)\to \underset{S+0.4}{{\mathrm{M}}^{+3}}\left(\mathrm{aq}\right)+\underset{3\mathrm{s}}{3{\mathrm{OH}}^{-}}\left(\mathrm{aq}\right)$

And the solubility of M(OH)₃ will get suppressed in presence of 0.4 M MX₃ solution.

${\mathrm{MX}}_3\to {\mathrm{M}}_{0.4+{\mathrm{S}}^{\text{'}}}^{+3}\left(\mathrm{aq}\right)+3{\mathrm{X}}_{3\times 0.4}^{-}\left(\mathrm{aq}\right)$

the solubility of M(OH)₃ is suppressed, S' will be negligible. i.e. 0.4 + S' ~ 0.4

K_sp = (S' + 0.4)  x (3S')³

⇒  K_sp = 0.4 x S'³ x 27

$\Rightarrow \frac{{\mathrm{K}}_{\mathrm{sp}}}{10.8}={\left({\mathrm{S}}^{\text{'}}\right)}^3$

$\Rightarrow S^{\text{'}}={\left(\frac{K_{sp}}{10.8}\right)}^{\frac{1}{3}}$