Show that $2^n+3^n+5^n+6^n$ isn't a perfect cube, for all $n \in \mathbb{N}$.

Question

Prove that number $A=2^n+3^n+5^n+6^n$ is not a perfect cube, for all $n \in \mathbb{N}$.

I managed to prove that $$2^n+3^n=5^n \mod 6$$ I also tried some particular cases.

Answer 1

By noting that

$$ (3q+r)^3 = 9(3q^3+3q^2r+qr^2) + r^3 \equiv r^3 \pmod{9}, $$

we get

$$ n^3 \equiv 0, 1, 8 \pmod{9}, \qquad \forall n \in \mathbb{Z}. $$

On the other hand, we can check that

$$ 2^n + 3^n + 5^n + 6^n \equiv 2, 7 \pmod{9}, \qquad \forall n \geq 1. $$