If $x=720$ and $ax$ is a perfect cube where $a$ is a positive integer, what is the smallest possible value of $\sqrt[3]{ax}$?
To find the smallest possible value of $\sqrt[3]{ax}$, I need to minimize $a$ such that $ax$ is a perfect cube. I can start by factoring $x=720$ into prime factors: $720=2^4\cdot 3^2\cdot 5$. Now I need to find the smallest $a$ that will make the exponents of each prime factor in $ax$ a multiple of 3. For 2, the exponent is 4, so I need to multiply by 2^2 to make it 6. For 3, the exponent is 2, so I need to multiply by 3 to make it 3. For 5, the exponent is 1, so I need to multiply by 5^2 to make it 3. Therefore, the smallest $a$ is $a=2^2\cdot 3\cdot 5^2=300$. This means that $ax=300\cdot 720=2^6\cdot 3^3\cdot 5^3$. Taking the cube root, I get $\sqrt[3]{ax}=2^2\cdot 3\cdot 5=60$. 