Let $n$ be a positive integer. Prove that if $18|n^3$, then $18|n^2$. Let $n$ be a positive integer. Prove that if $18|n^3$, then $18|n^2$.
I realize that this does not also apply to $18|n$, since it works for $n=6$. I think that in order to solve this, the fundamental theorem of arithmetic needs to be used. 

Suppose $18$ divides $n^3$. This means that there is some integer $k$ such that $18k=n^3$. Notice that $18=2\cdot 3^2$. So $2$ divides $n^3$ and $3^2$ divides $n^3$. 
If $2$ divides $n^3$, what can you say about $n$ itself?
If $3$ divides $n^3$, again what can be said about $n$?
Consider prime factorizations. 
I hope this helps!
