We need to prove that $abc$ is divisible by $2$ and by $3$. We will give proofs by contradiction.

Suppose $abc$ odd. This implies that $a$, $b$ and $c$ are odd. Therefore, $a^3 + b^3 + c^3$ is odd and certainly not divisible by $18$. This contradiction shows that $abc$ is even.

Suppose that $abc$ is not divisible by $3$. Then $a$, $b$ and $c$ are not divisible by $3$, i.e. they are in (possibly distinct) congruence classes among the following congruence classes mod $9$.

We conclude that $a^3 + b^3 + c^3$ is equal to $-3$, $-1$, $1$ or $3 \pmod 9$. Therefore, $a^3 + b^3 + c^3$ is not divisible by $9$ and consequently not by $18$. This contradiction shows that $abc$ is divisible by $3$.