Mathematical Reflections 2016, Issue 5 - Problem O387

Problem:
Are there integers $n$ for which $$3^{6n-3}+3^{3n-1}+1$$ is a perfect cube?

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Clearly if $n \leq 0$, the number $3^{6n-3}+3^{3n-1}+1$ is never a perfect cube. Let $n>0$. Then,
$$(3^{2n-1})^3=3^{6n-3} < 3^{6n-3}+3^{3n-1}+1 < 3^{6n-3}+3^{4n-1}+3^{2n}+1=(3^{2n-1}+1)^3.$$
So, there are no integers $n$ such that $3^{6n-3}+3^{3n-1}+1$ is a perfect cube.