Prove that if $9\mid(x^3 + y^3 + z^3)$ then $3\mid xyz$ for integers $x, y, z$

Question

how would I prove that if 9 divides $x^3+y^3+z^3$ then $3$ divides $xyz$?

I've thought about it and don't really know where to start. So far, I know that $x^3+y^3+z^3$ is congruent to $0 (\mod 9)$.

I don't know where to go from here.

Answer 1

Observe that for any integer $a$ we have $a^3 \equiv 0, \pm 1 \pmod{9}$. So for $x^3+y^3+z^3 \equiv 0 \pmod{9}$ we can only have the following scenarios:

1. Each of $x^3, y^3$ and $z^3$ are $0 \bmod{9}$.
2. One of them is $0 \bmod{9}$ and the remaining two are $1 \bmod{9}$ and $-1 \bmod{9}$.

But in both cases we have at least one (say) $x^3 \equiv 0 \pmod{9}$. This implies $x^3 \equiv 0 \pmod{3}$. Consequently $x \equiv 0 \pmod{3}$. This implies $xyz \equiv 0 \pmod{3}$.