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.
 A: At first, work $\pmod 3$.  Clearly we can't have either $a,b\equiv 0 \pmod 3$ so by symmetry we only have to consider $3$ pairs $$(a,b)\in \{(1,1),(1,2), (2,2)\}$$
A quick computation shows that only $(2,2)$ works so we must have $a=3m+2,b=3n+2$ for some integers $m,n$.
We now check that $$(3m+2)(3n+2)(3m+3n+4)+2\equiv 0\pmod 9$$ and we are done.
May be worth noting:  to do the final check you don't need to multiply everything out (though that's not all that hard).  It's clear that the coefficients of $n^2,m^2,mn$ are divisible by $9$.  The coefficients of  $m,n$ are both $3\times 2\times 4+2\times 2 \times 3=24+12=36\equiv 0 \pmod 9$ so the product is $16\pmod 9$ and we are done (since adding $2$ to the product gives us $18$ which of course is $0\pmod 9$).
Note:  I believe this is substantially similar to the argument you give in your post, but your version appears to be more complex and involved.  I think the way I've written it gets at the main issues quickly.
A: 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 $9 \mid [ab(a + b) + 2]$.
A: Clearly $3\nmid a,b$. Fermat little theorem we have $a^3\equiv_3 a$ so multiplying given relation with $a$ on the right we get $$3\mid a^2b^2+ab+2a\implies 3\mid 4a^2b^2+4ab+8a$$
$$\implies 3\mid (2ab+1)^2+8a-1\implies a+1\equiv_3 (2ab+1)^2$$ 
Since $(2ab+1)^2 \equiv_3 0,1$ we see, since $3\nmid a$ that $a\equiv_3 -1$. The same is true for $b\equiv_3 -1$. So $a=3x-1$ and $b=3y-1$. Now we have:
$$ (3x-1)^2(3y-1)+(3y-1)^2(3x-1)+2\equiv_9$$ $$\equiv_9(3x+1)(3y-1)+(3y+1)(3x-1)+2\equiv_9  0$$
