Find all $n\in\mathbb N$ such that by removing the last $3$ digits of $n$, $\sqrt[3]{n}$ is obtained. 
Find all $n\in\mathbb N$ such that by removing the last $3$ digits of $n$, $\sqrt[3]{n}$ is obtained.

I found that $n=32768$ is a solution. Is there any other of $n$? It'll be best if no computer is used. Thank you.

Focus instead on the cube root $m$. It must satisfy
$$1000m\le m^3<1000(m+1).$$
From the first inequality, get $m^2\ge 1000$, i.e., $m\ge32$. Notice that $m=33$ violates the second inequality ($33^3=35937>34000$), and this gets worse with larger $m$. So $m=32$ is the only solution.
