Calculate the molar solubility (S) of a salt like zirconium phosphate of molecular formula ${\left({\mathrm{Zr}}^{ }\right)}_3{\left({\mathrm{PO}}_4^{ }\right)}_4$

Consider a salt, zirconium phosphate of molecular formula ${\left({\mathrm{Zr}}^{4+}\right)}_3{\left({\mathrm{PO}}_4^{3-}\right)}_4$. It dissociates into 3 zirconium cations of charge +4 and 4 phosphate anions of charge -3.
$\left[{\mathrm{Zr}}^{4+}\right]=3\mathrm{S} \text{ and } \left[{\mathrm{PO}}_4^{3-}\right]=4\mathrm{S}$

and     ${\mathrm{K}}_{\mathrm{sp}}=\left(3\mathrm{S}{)}^3\right(4\mathrm{S}{)}^4=6912(\mathrm{S}{)}^7$

or      $\mathrm{S}={\left\{{\mathrm{K}}_{\mathrm{sp}}/\left(3^3\times 4^4\right)\right\}}^{1/7}={\left({\mathrm{K}}_{\mathrm{sp}}/6912\right)}^{\mathrm{I}/7}$