The polynomial is (almost certainly) quadratic and so cannot be always divisible even by $3$, as this would force divisibility by the cubic $x(x-1)(x-2)$. The other option is for it to be identically $0$ mod $3$, i.e. $a+b+c, ab + bc + ca$ and $abc - 1$ to be divisible by $3$ and now it's just a simple case bash.