For which real numbers $a$, there exist pairwise different real numbers $x, y, z$ satisfying
$$\frac{x^3+a}{y+z}=\frac{y^3+a}{x+z}=\frac{z^3+a}{x+y}= -3.$$
1. Given the equations:
   \[
   \frac{x^3 + a}{y + z} = \frac{y^3 + a}{x + z} = \frac{z^3 + a}{x + y} = -3
   \]
   we can rewrite them as:
   \[
   x^3 + a = -3(y + z), \quad y^3 + a = -3(x + z), \quad z^3 + a = -3(x + y)
   \]

2. Subtracting the second equation from the first, we get:
   \[
   x^3 - y^3 = -3(y + z) + 3(x + z) \implies x^3 - y^3 = 3(x - y)
   \]
   Factoring the left-hand side, we have:
   \[
   (x - y)(x^2 + xy + y^2) = 3(x - y)
   \]
   Since \(x \neq y\), we can divide both sides by \(x - y\):
   \[
   x^2 + xy + y^2 = 3
   \]

3. Similarly, from the other pairs of equations, we get:
   \[
   y^2 + yz + z^2 = 3
   \]
   \[
   z^2 + zx + x^2 = 3
   \]

4. Adding these three equations, we obtain:
   \[
   (x^2 + xy + y^2) + (y^2 + yz + z^2) + (z^2 + zx + x^2) = 9
   \]
   Simplifying, we get:
   \[
   2(x^2 + y^2 + z^2) + (xy + yz + zx) = 9
   \]

5. Since \(x + y + z = 0\) (from the condition \((x - z)(x + z + y) = 0\)), we have:
   \[
   (x + y + z)^2 = 0 \implies x^2 + y^2 + z^2 + 2(xy + yz + zx) = 0
   \]
   Substituting \(x^2 + y^2 + z^2 + 2(xy + yz + zx) = 0\) into the previous equation, we get:
   \[
   2(x^2 + y^2 + z^2) + (xy + yz + zx) = 9
   \]
   \[
   2(x^2 + y^2 + z^2) - (x^2 + y^2 + z^2) = 9
   \]
   \[
   x^2 + y^2 + z^2 = 9
   \]

6. Now, we need to find the range of \(a\). From the original equations, we have:
   \[
   x^3 + a = -3(y + z)
   \]
   Since \(x + y + z = 0\), we have \(y + z = -x\). Substituting this, we get:
   \[
   x^3 + a = 3x \implies a = 3x - x^3
   \]

7. To find the range of \(a\), we need to find the range of \(3x - x^3\) for \(x \in [-2, 2]\). The function \(f(x) = 3x - x^3\) is a cubic polynomial. We find its critical points by differentiating:
   \[
   f'(x) = 3 - 3x^2
   \]
   Setting \(f'(x) = 0\), we get:
   \[
   3 - 3x^2 = 0 \implies x^2 = 1 \implies x = \pm 1
   \]

8. Evaluating \(f(x)\) at the critical points and endpoints:
   \[
   f(-2) = 3(-2) - (-2)^3 = -6 + 8 = 2
   \]
   \[
   f(-1) = 3(-1) - (-1)^3 = -3 + 1 = -2
   \]
   \[
   f(1) = 3(1) - 1^3 = 3 - 1 = 2
   \]
   \[
   f(2) = 3(2) - 2^3 = 6 - 8 = -2
   \]

9. Therefore, the range of \(a\) is \((-2, 2)\), excluding 0 because \(x, y, z\) are pairwise different.

The final answer is \( \boxed{ a \in (-2, 2) \setminus \{0\} } \)