Find all triples $ (x,y,z)$ of real numbers which satisfy the simultaneous equations

\[ x \equal{} y^3 \plus{} y \minus{} 8\]

\[y \equal{} z^3 \plus{} z \minus{} 8\]

\[ z \equal{} x^3 \plus{} x \minus{} 8.\]
1. We start with the given system of equations:
   \[
   x = y^3 + y - 8
   \]
   \[
   y = z^3 + z - 8
   \]
   \[
   z = x^3 + x - 8
   \]

2. First, we check if \((x, y, z) = (2, 2, 2)\) is a solution:
   \[
   x = 2, \quad y = 2, \quad z = 2
   \]
   Substituting \(y = 2\) into the first equation:
   \[
   x = 2^3 + 2 - 8 = 8 + 2 - 8 = 2
   \]
   Substituting \(z = 2\) into the second equation:
   \[
   y = 2^3 + 2 - 8 = 8 + 2 - 8 = 2
   \]
   Substituting \(x = 2\) into the third equation:
   \[
   z = 2^3 + 2 - 8 = 8 + 2 - 8 = 2
   \]
   Therefore, \((2, 2, 2)\) is indeed a solution.

3. Next, we assume \((x - 2)(y - 2)(z - 2) \neq 0\) and derive a contradiction:
   \[
   x - 2 = (y - 2)((y + 1)^2 + 4)
   \]
   \[
   y - 2 = (z - 2)((z + 1)^2 + 4)
   \]
   \[
   z - 2 = (x - 2)((x + 1)^2 + 4)
   \]

4. Multiplying these three equations:
   \[
   (x - 2)(y - 2)(z - 2) = (y - 2)(z - 2)(x - 2)((y + 1)^2 + 4)((z + 1)^2 + 4)((x + 1)^2 + 4)
   \]
   Dividing both sides by \((x - 2)(y - 2)(z - 2)\) (which is non-zero by assumption):
   \[
   1 = ((y + 1)^2 + 4)((z + 1)^2 + 4)((x + 1)^2 + 4)
   \]

5. Notice that \((y + 1)^2 + 4 \geq 4\), \((z + 1)^2 + 4 \geq 4\), and \((x + 1)^2 + 4 \geq 4\). Therefore:
   \[
   ((y + 1)^2 + 4)((z + 1)^2 + 4)((x + 1)^2 + 4) \geq 4 \times 4 \times 4 = 64
   \]
   This implies:
   \[
   1 \geq 64
   \]
   which is a contradiction.

6. Since the assumption \((x - 2)(y - 2)(z - 2) \neq 0\) leads to a contradiction, the only possible solution is when \((x - 2)(y - 2)(z - 2) = 0\). Given that \((2, 2, 2)\) is a solution, we conclude that it is the only solution.

The final answer is \( \boxed{ (x, y, z) = (2, 2, 2) } \).