Let $a \circ b=a+b-a b$. Find all triples $(x, y, z)$ of integers such that $(x \circ y) \circ z+(y \circ z) \circ x+(z \circ x) \circ y=0$.
Note that

$$
(x \circ y) \circ z=x+y+z-x y-y z-x z+x y z=(x-1)(y-1)(z-1)+1 .
$$

Hence

$$
(x \circ y) \circ z+(y \circ z) \circ x+(z \circ x) \circ y=3((x-1)(y-1)(z-1)+1) \text {. }
$$

Now, if the required equality holds we have $(x-1)(y-1)(z-1)=-1$. There are only four possible decompositions of -1 into a product of three integers. Thus we have four such triples, namely $(0,0,0)$, $(0,2,2),(2,0,2)$ and $(2,2,0)$.