Note that $a+ab+b+bc+c+ca=a+b+c \Rightarrow ab+bc+ca=0$ $\Rightarrow 3abc=a^2-ab+b^2-bc+c^2-ca=(a+b+c)^2-2(ab+bc+ca)=(a+b+c)^2$, but $3(a+1)(b+1)(c+1)=3abc+3ab+3bc+3ca+3a+3b+3c+3$ $=(a+b+c)(a+b+c+3)+3$ and $a(b+1)b(c+1)c(a+1)=cab\Rightarrow a+b+c=-3\Rightarrow abc=3$.

The system can be solved trigonometrically. Challenge: Show that the system has solutions $(0,0,0)$ and the cyclic permutations of $\left(-1+2\cos\frac{7\pi}{9},-1+2\cos\frac{5\pi}{9},-1+2\cos\frac{\pi}{9}\right)$. Of course you can verify the latter solutions obey $abc=3$, the answer to the Olympiad question.

I don't know how to use LATEx. Obviously ab+bc+ac=0,multiplying the three equations, we arrive at (a+1)(b+1)(c+1)=1,so,a+b+c=-abc.Multiplying each equation by the isolated term and adding,we have a² + b² + c²=3abc,but ab+bc+ac=0,so (a + b + c)²=3abc.then we can conclude that abc=3