Determine all real solutions $x, y, z$ of the following system of equations: $\begin{cases}
x^3 - 3x = 4 - y \\
2y^3 - 6y = 6 - z \\
3z^3 - 9z = 8 - x\end{cases}$
We are given the system of equations:
\[
\begin{cases}
x^3 - 3x = 4 - y \\
2y^3 - 6y = 6 - z \\
3z^3 - 9z = 8 - x
\end{cases}
\]

1. **Rewrite each equation in a more convenient form:**
   \[
   \begin{cases}
   x^3 - 3x + y = 4 \\
   2y^3 - 6y + z = 6 \\
   3z^3 - 9z + x = 8
   \end{cases}
   \]

2. **Isolate \( y \), \( z \), and \( x \) respectively:**
   \[
   \begin{cases}
   y = 4 - x^3 + 3x \\
   z = 6 - 2y^3 + 6y \\
   x = 8 - 3z^3 + 9z
   \end{cases}
   \]

3. **Substitute \( y \) from the first equation into the second equation:**
   \[
   z = 6 - 2(4 - x^3 + 3x)^3 + 6(4 - x^3 + 3x)
   \]

4. **Substitute \( z \) from the second equation into the third equation:**
   \[
   x = 8 - 3(6 - 2(4 - x^3 + 3x)^3 + 6(4 - x^3 + 3x))^3 + 9(6 - 2(4 - x^3 + 3x)^3 + 6(4 - x^3 + 3x))
   \]

5. **Check for simple solutions by trial:**
   - Let \( x = 2 \):
     \[
     y = 4 - 2^3 + 3 \cdot 2 = 4 - 8 + 6 = 2
     \]
     \[
     z = 6 - 2 \cdot 2^3 + 6 \cdot 2 = 6 - 16 + 12 = 2
     \]
     \[
     x = 8 - 3 \cdot 2^3 + 9 \cdot 2 = 8 - 24 + 18 = 2
     \]
     Thus, \( x = y = z = 2 \) is a solution.

6. **Check for other potential solutions:**
   - Consider the product of the transformed equations:
     \[
     (x-2)(x+1)^2 \cdot 2(y-2)(y+1)^2 \cdot 3(z-2)(z+1)^2 = (2-y)(2-z)(2-x)
     \]
     Simplifying, we get:
     \[
     6(x+1)^2(y+1)^2(z+1)^2 = -1
     \]
     Since the left-hand side is always non-negative (as it is a product of squares), it cannot equal \(-1\). Therefore, there are no other real solutions.

Conclusion:
The only real solution to the system is \( x = y = z = 2 \).

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