Drop a spoon of silver chloride into water and almost nothing dissolves — but not NOTHING. A tiny saturated equilibrium sets up between the solid and its ions, and because the solid is a pure solid it drops out of the expression, leaving only the ions. That leftover is the solubility product $K_{sp}$. Two symbols do all the work here: $K_{sp}$ is the solubility product, and $s$ is the molar solubility — how many moles of the salt dissolve per litre before the solution gives up, in mol/L.

The shape of the salt sets the algebra. A 1:1 salt like AgCl releases one cation and one anion, so dissolving $s$ moles gives $s$ of each: $K_{sp} = (s)(s) = s^{2}$, and running it backwards, $s = \sqrt{K_{sp}}$. An AB₂ or A₂B salt like CaF₂ releases three ions — two of one, one of the other — so dissolving $s$ moles gives $2s$ of the doubled ion and $s$ of the other: $K_{sp} = (2s)^{2}(s) = 4s^{3}$. That is where the 4 comes from, and it is squared into existence, not decoration. Backwards: $s = \sqrt[3]{\dfrac{K_{sp}}{4}}$ — divide the 4 away first, then take the root.

Which sets up the trap this lesson exists for. A 1:1 product is $(\mathrm{mol/L})^{2}$; a three-ion product is $(\mathrm{mol/L})^{3}$. They are printed in the same column of the same table with no units beside them, and they are not the same quantity. Comparing them directly is ranking an area against a volume. A salt with the SMALLER $K_{sp}$ can be the more soluble of the two, and you will meet one before this lesson is over.