jmc

First we find the prime factorization of 720, which is $2^4\cdot 3^2\cdot 5$. In order to make a perfect cube, we need two more factors of 2, another factor of 3 and two more factors of 5. So if $a=2^2\cdot 3\cdot 5^2$, we have $ax=(2^2\cdot 3\cdot 5^2)(2^4\cdot 3^2\cdot 5)=2^6\cdot 3^3\cdot 5^3$. That means $\sqrt[3]{ax}=2^2\cdot 3\cdot 5=\boxed{60}$.