Find all real solutions of the system of equations $\begin{cases} 
(x + y) ^3 = z \\ (y + z) ^3 = x \\ ( z+ x) ^3 = y \end{cases} $

(Based on an idea by A . Aho , J. Hop croft , J. Ullman )
1. We start with the given system of equations:
   \[
   \begin{cases} 
   (x + y)^3 = z \\
   (y + z)^3 = x \\
   (z + x)^3 = y 
   \end{cases}
   \]

2. Subtract the first equation from the third equation:
   \[
   (z + x)^3 - (x + y)^3 = y - z
   \]
   Using the identity for the difference of cubes, we get:
   \[
   (z + x - (x + y)) \left[(z + x)^2 + (z + x)(x + y) + (x + y)^2\right] = y - z
   \]
   Simplifying the first term:
   \[
   (z - y) \left[(z + x)^2 + (z + x)(x + y) + (x + y)^2\right] = y - z
   \]
   Since \(z - y = -(y - z)\), we can rewrite the equation as:
   \[
   (y - z) \left[(z + x)^2 + (z + x)(x + y) + (x + y)^2\right] = z - y
   \]
   This implies:
   \[
   (y - z) \left[(z + x)^2 + (z + x)(x + y) + (x + y)^2 + 1\right] = 0
   \]
   Therefore, we have:
   \[
   y - z = 0 \quad \text{or} \quad \left[(z + x)^2 + (z + x)(x + y) + (x + y)^2 + 1\right] = 0
   \]
   Since the second term is always positive, we must have:
   \[
   y = z
   \]

3. Similarly, subtract the third equation from the second equation:
   \[
   (y + z)^3 - (z + x)^3 = x - y
   \]
   Using the same identity for the difference of cubes, we get:
   \[
   (y - x) \left[(y + z)^2 + (y + z)(z + x) + (z + x)^2\right] = x - y
   \]
   Simplifying the first term:
   \[
   (y - x) \left[(y + z)^2 + (y + z)(z + x) + (z + x)^2 + 1\right] = 0
   \]
   Therefore, we have:
   \[
   y - x = 0 \quad \text{or} \quad \left[(y + z)^2 + (y + z)(z + x) + (z + x)^2 + 1\right] = 0
   \]
   Since the second term is always positive, we must have:
   \[
   y = x
   \]

4. From steps 2 and 3, we have:
   \[
   x = y = z
   \]

5. Substitute \(x = y = z\) into the first equation:
   \[
   (x + x)^3 = x \implies (2x)^3 = x \implies 8x^3 = x
   \]
   Solving for \(x\), we get:
   \[
   8x^3 - x = 0 \implies x(8x^2 - 1) = 0
   \]
   Therefore:
   \[
   x = 0 \quad \text{or} \quad 8x^2 = 1 \implies x^2 = \frac{1}{8} \implies x = \pm \frac{\sqrt{2}}{4}
   \]

6. Thus, the solutions are:
   \[
   x = y = z = 0 \quad \text{or} \quad x = y = z = \pm \frac{\sqrt{2}}{4}
   \]

The final answer is \( \boxed{ x = y = z = 0 } \) or \(x = y = z = \pm \frac{\sqrt{2}}{4}\)