Divisibility of three polynomial terms

Question

So here is the statement that im having trouble proving:

If $9\mid x^3+y^3+z^3$ then $3\mid xyz$ for integers $x,y,z$.

I tried applying the definition of divisibility but that doesn't seem to lead me anywhere. I also tried applying the definition of congruence to no avail.

Answer 1

HINT : Note that for $a\in\mathbb Z$,$$a^3\equiv 0,\pm 1\pmod 9.$$

Answer 2

There are three cubes mod 9, they are 0, 1, and -1. In order for $9\mid x^3+y^3+z^3$, i.e., that $x^3+y^3+z^3\equiv 0\pmod 9$ it must be that one of the terms is equivalent to $0\pmod 9$. Assume that it is $x^3$. Since $9|x^3$, $3|x$.