18.1. What is the smallest natural number by which 720 must be multiplied so that the result is a cube of a natural number?
18.1. We have: $720=2^{4} \cdot 3^{2} \cdot 5$. To obtain a perfect cube from this number, it needs to be multiplied by a number such that the exponents of two, three, and five in the resulting product are divisible by three. Clearly, the smallest such multiplier will be the number $2^{2} \cdot 3 \cdot 5^{2}=300$. Indeed, in this case $720 \cdot 300=$ $=2^{6} \cdot 3^{3} \cdot 5^{3}=\left(2^{2} \cdot 3 \cdot 5\right)^{3}=60^{3}$.