When I noticed that $3^3+4^3+5^3=6^3$, I wondered if there are any other times where $(a-1)^3+a^3+(a+1)^3$ equals another cube. That expression simplifies to $3a(a^2+2)$ and I'm still trying to find another value of $a$ that satisfies the condition (the only one found is $a=4$)
Is this impossible? (It doesn't happen for $3 \leq a \leq 10000$) Is it possible to prove?