We compute $2^{8}+1=257$. We\textquotesingle re all familiar with what $6^{3}$ is, namely, which is too small. The smallest cube greater than it is $7^{3}=343$. $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^{6})^{3}=64^{3}$, which therefore clearly will be the largest cube less than $2^{18}+1$. Therefore, the required number of cubes is $64-7+1=58$. 
How many perfect cubes lie between $2^{8}+1$ and $2^{18}+1$, inclusive? (2018 AMC 8 Problem, Question \#25) 
(a) $$4$$ 
(b) $$9$$ 
(c) $$10$$ 
(d) $$57$$ 
(e) $$58$$ 