So $ n^3 \equiv 0 \pmod{8} $, which implies $ n \equiv 0 \pmod{2} $, and in fact $ n $ must be divisible by $2$. But to get $n^3 \equiv 0 \pmod{8}$, $n$ must be divisible by $2$, and checking small cubes mod $8$: $0^3 = 0$, $2^3 = 8 \equiv 0$, $4^3 = 64 \equiv 0$, $6^3 = 216 \equiv 0$. So all even $n$ satisfy $n^3 \equiv 0 \pmod{8}$. So $ n \equiv 0 \pmod{2} $ is sufficient,