Can $36\underbrace{11...1}_{n \text { times}}$ be a perfect cube?
Inspired from the question What is the next perfect square of the form 14444... in decimal notation? ,$361$ is the only perfect square that is form of $361…1$, since the others is $3\pmod{4}$, but can $36\underbrace{11…1}_{n \text{ times}}$ be a perfect cube when $n\geq2$?
I tried to use Pari GP and checked the values of $n\leq10^4$ manually, but none of the numbers shown is a perfect cube.
Hint: $36\underbrace{11…1}_{n \text{ times}}\equiv3\pmod{4}$ when $n\ge2$ and $36\underbrace{11…1}_{n \text{ times}}\equiv0,\pm1\pmod{9}$ when $n\equiv0,\pm1\pmod{9}$.