If $x^3 - y^3 = 208$ and $x - y = 4$, $xy = ?$

Using $x^3-y^3 = (x-y)^3 + 3xy(x-y)$, substitute to get $208 = 4^3 + 3xy(4) = 64 + 12xy$. Solving, $12xy = 144$, so $xy = 12$.