Solve for integers $x,y,z$: \[ \{ \begin{array}{ccc} x + y &=& 1 - z \\ x^3 + y^3 &=& 1 - z^2 . \end{array}  \]
We are given the system of equations:
\[
\begin{cases}
x + y = 1 - z \\
x^3 + y^3 = 1 - z^2
\end{cases}
\]

1. **Substitute \( z = 1 - x - y \) into the second equation:**

   \[
   x^3 + y^3 = 1 - (1 - x - y)^2
   \]

2. **Simplify the right-hand side:**

   \[
   1 - (1 - x - y)^2 = 1 - (1 - 2(x + y) + (x + y)^2) = 1 - (1 - 2(x + y) + (x + y)^2)
   \]

   \[
   = 1 - 1 + 2(x + y) - (x + y)^2 = 2(x + y) - (x + y)^2
   \]

3. **Rewrite the equation:**

   \[
   x^3 + y^3 = 2(x + y) - (x + y)^2
   \]

4. **Use the identity for the sum of cubes:**

   \[
   x^3 + y^3 = (x + y)(x^2 - xy + y^2)
   \]

   Substituting \( x + y = 1 - z \):

   \[
   (1 - z)(x^2 - xy + y^2) = 2(1 - z) - (1 - z)^2
   \]

5. **Simplify the right-hand side:**

   \[
   2(1 - z) - (1 - z)^2 = 2 - 2z - (1 - 2z + z^2) = 2 - 2z - 1 + 2z - z^2 = 1 - z^2
   \]

   Thus, we have:

   \[
   (1 - z)(x^2 - xy + y^2) = 1 - z^2
   \]

6. **Case 1: \( 1 - z = 0 \)**

   \[
   z = 1
   \]

   Substituting \( z = 1 \) into the first equation:

   \[
   x + y = 1 - 1 = 0 \implies y = -x
   \]

   Substituting \( y = -x \) into the second equation:

   \[
   x^3 + (-x)^3 = 1 - 1^2 \implies x^3 - x^3 = 0
   \]

   This is always true, so we have the solutions:

   \[
   (x, y, z) = (k, -k, 1) \quad \text{for all } k \in \mathbb{Z}
   \]

7. **Case 2: \( 1 - z \neq 0 \)**

   \[
   x^2 - xy + y^2 = 1 + z
   \]

   Substituting \( x + y = 1 - z \):

   \[
   x^2 + 2xy + y^2 = (1 - z)^2
   \]

   \[
   (x + y)^2 = 1 - 2z + z^2
   \]

   \[
   (1 - z)^2 = 1 - 2z + z^2
   \]

   Substituting \( x^2 - xy + y^2 = 1 + z \):

   \[
   x^2 - xy + y^2 = 1 + z
   \]

   \[
   x^2 + 2xy + y^2 = 1 + z + 2xy
   \]

   \[
   (x + y)^2 = 1 + z + 2xy
   \]

   \[
   (1 - z)^2 = 1 + z + 2xy
   \]

   \[
   1 - 2z + z^2 = 1 + z + 2xy
   \]

   \[
   z^2 - 3z - 2xy = 0
   \]

   Solving for \( y \):

   \[
   y = \frac{z(z - 3)}{3x}
   \]

   Substituting into the quadratic equation:

   \[
   3x^2 + (3z - 3)x + z^2 - 3z = 0
   \]

   The discriminant must be a perfect square for \( x \) to be an integer:

   \[
   \Delta = (3z - 3)^2 - 4 \cdot 3 \cdot (z^2 - 3z) = 9z^2 - 18z + 9 - 12z^2 + 36z = -3z^2 + 18z + 9
   \]

   \[
   \Delta = -3(z^2 - 6z - 3)
   \]

   For the discriminant to be a perfect square, \( z \) must be an integer such that \( -3(z^2 - 6z - 3) \) is a perfect square. Checking values, we find that \( z = 0, 3, 6 \) work.

   \[
   \boxed{(k, -k, 1), (0, 1, 0), (1, 0, 0), (0, -2, 3), (-2, 0, 3), (-2, -3, 6), (-3, -2, 6)}
   \]