China Hong Kong Mathematical Olympiad

The solutions are $(a, b, c) = (1, 0, 0), (2, -1, -1)$ up to permutation.
Let $s = a + b + c$. Consider
$$P(x) = (x - a)(x - b)(x - c) = x^3 - sx^2 + (ab + bc + ca)x - abc.$$
Putting $x = s$, we obtain $(b+c)(c+a)(a+b) = (ab+bc+ca)s - abc$. Therefore, the given equation becomes
$$(ab + bc + ca)s - abc + 2s^3 = 2 - 2abc.$$
This is the same as $-s^3 - s^3 - (ab + bc + ca)s - abc = -2$, i.e. $P(-s) = -2$. In other words, we have
$$(2a + b + c)(a + 2b + c)(a + b + 2c) = 2.$$
(One can prove this by direct expansion.) By symmetry, it suffices to consider a few cases as follows.
* If $2a + b + c = 2$ and $a + 2b + c = a + b + 2c = 1$, adding these we obtain $4(a + b + c) = 4$. So we have $a + b + c = 1$, and hence $(a, b, c) = (1, 0, 0)$.
* If $2a + b + c = 2$ and $a + 2b + c = a + b + 2c = -1$, adding these we obtain $4(a + b + c) = 0$. So we have $a + b + c = 0$, and hence $(a, b, c) = (2, -1, -1)$.
* If $2a + b + c = -2$, $a + 2b + c = \pm 1$ and $a + b + 2c = \mp 1$, adding these we obtain $4(a + b + c) = -2$. There is no integer solution.
This shows the only solutions are $(a, b, c) = (1, 0, 0), (2, -1, -1)$ up to permutation.