Let $a\circ b=a+b-ab$. Find all triples $(x,y,z)$ of integers such that
\[(x\circ y)\circ z +(y\circ z)\circ x +(z\circ x)\circ y=0\]
1. Given the operation \( a \circ b = a + b - ab \), we can rewrite it as:
   \[
   1 - a \circ b = 1 - (a + b - ab) = 1 - a - b + ab = (1 - a)(1 - b)
   \]
   This shows that the operation \( \circ \) is isomorphic to regular multiplication under the transformation \( x \rightarrow 1 - x \).

2. Define the transformation \( \phi(a) = 1 - a \). Notice that:
   \[
   \phi(\phi(a)) = \phi(1 - a) = 1 - (1 - a) = a
   \]
   This means \( \phi \) is an involution (applying it twice returns the original value).

3. Using the transformation, we can express the operation \( \circ \) in terms of \( \phi \):
   \[
   \phi(a \circ b) = \phi(a + b - ab) = 1 - (a + b - ab) = (1 - a)(1 - b) = \phi(a) \cdot \phi(b)
   \]
   Therefore, \( \phi(a \circ b) = \phi(a) \cdot \phi(b) \).

4. Now, consider the expression \( (a \circ b) \circ c \):
   \[
   (a \circ b) \circ c = \phi(\phi(a) \cdot \phi(b)) \circ \phi(c) = \phi(\phi(a) \cdot \phi(b) \cdot \phi(c))
   \]
   Since \( \phi(\phi(a)) = a \), we have:
   \[
   (a \circ b) \circ c = \phi(\phi(a) \cdot \phi(b) \cdot \phi(c)) = 1 - (1 - a)(1 - b)(1 - c)
   \]

5. Simplify the expression:
   \[
   (a \circ b) \circ c = 1 - (1 - a)(1 - b)(1 - c) = 1 - (1 - a - b + ab)(1 - c) = 1 - (1 - a - b + ab - c + ac + bc - abc)
   \]
   \[
   = a + b + c - ab - ac - bc + abc
   \]

6. Given the problem's condition:
   \[
   (x \circ y) \circ z + (y \circ z) \circ x + (z \circ x) \circ y = 0
   \]
   Using the simplified form:
   \[
   (x \circ y) \circ z = (x-1)(y-1)(z-1) + 1
   \]
   Similarly:
   \[
   (y \circ z) \circ x = (y-1)(z-1)(x-1) + 1
   \]
   \[
   (z \circ x) \circ y = (z-1)(x-1)(y-1) + 1
   \]

7. Summing these expressions:
   \[
   (x-1)(y-1)(z-1) + 1 + (y-1)(z-1)(x-1) + 1 + (z-1)(x-1)(y-1) + 1 = 0
   \]
   \[
   3((x-1)(y-1)(z-1)) + 3 = 0
   \]
   \[
   3((x-1)(y-1)(z-1) + 1) = 0
   \]
   \[
   (x-1)(y-1)(z-1) = -1
   \]

8. We need three integers whose product is \(-1\). The integers must be either \(1\) or \(-1\), and the number of \(-1\)'s must be odd. The possible combinations are:
   - One \(-1\) and two \(1\)'s: \((x-1, y-1, z-1) = (-1, 1, 1)\) or permutations.
   - Three \(-1\)'s: \((x-1, y-1, z-1) = (-1, -1, -1)\).

9. Solving for \(x, y, z\):
   - For \((x-1, y-1, z-1) = (-1, 1, 1)\), we get \((x, y, z) = (0, 2, 2)\) and permutations.
   - For \((x-1, y-1, z-1) = (-1, -1, -1)\), we get \((x, y, z) = (0, 0, 0)\).

Conclusion:
The only possible triples \((x, y, z)\) are:
\[
(x, y, z) = (0, 0, 0), (0, 2, 2), (2, 0, 2), (2, 2, 0)
\]

The final answer is \( \boxed{ (0, 0, 0), (0, 2, 2), (2, 0, 2), (2, 2, 0) } \).