Given that $a$ and $b$ are integers satisfied $3 \mid ab(a + b) + 2$, prove that $9 \mid ab(a + b) + 2$. 
Given that $a$ and $b$ are integers satisfy $3 \mid [ab(a + b) + 2]$, prove that $9 \mid [ab(a + b) + 2]$.

This problem is adapted from a recent competition. (The last few ones were and the next few ones will be.) I want to ask if the solution I have provided below is correct and if there are any other solutions which are more suitable more a test setting.

We have that $$3\mid ab(a + b) + 2 \implies ab(a + b) \equiv 1\text{ (mod 3)}$$
$$\implies \left[ \begin{align} ab \equiv 1&\text{ (mod 3) and } a + b \equiv 1\text{ (mod 3)}\\ ab \equiv -1&\text{ (mod 3) and } a + b \equiv -1\text{ (mod 3)} \end{align} \right.$$
If $ab \equiv -1\text{ (mod 3)}$ then we have that $\left[ \begin{align} a \equiv 1&\text{ (mod 3) and } b \equiv -1\text{ (mod 3)}\\ a \equiv -1&\text{ (mod 3) and } b \equiv 1\text{ (mod 3)} \end{align} \right. \implies 3 | a + b$
(which is contradictory to $a + b \equiv 2\text{ (mod 3)}$).
$\implies ab \equiv 1\text{ (mod 3)}$ and $a + b \equiv 1\text{ (mod 3)} \implies a \equiv b \equiv 2\text{ (mod 3)}$.
Let $a = 3m - 1$ and $b = 3n - 1 \ (m, n \in \mathbb Z)$.
We have that $$ab(a + b) + 2 = (3m - 1)(3n - 1)[(3m - 1) + (3n - 1)] + 2$$
$$= [9mn - 3(m + n) + 1][3(m + n) - 2] + 2$$
$$= [3(m + n) - 1][9mn - 3(m + n) + 2] - 9mn + 2$$
Furthermore, $$[3(m + n) - 1][9mn - 3(m + n) + 2] + 2 = 9(3m^2n + 3mn^2 - m^2 - n^2 - 3mn + m + n)$$
which is divisible by $9$. That means $$[3(m + n) - 1][9mn - 3(m + n) + 2] - 9mn + 2$$
divisible by $9$ or $9 \mid [ab(a + b) + 2]$.
