For real numbers $x$, $y$ and $z$, solve the system of equations:
$$x^3+y^3=3y+3z+4$$ $$y^3+z^3=3z+3x+4$$ $$x^3+z^3=3x+3y+4$$
To solve the system of equations:
\[
\begin{cases}
x^3 + y^3 = 3y + 3z + 4 \\
y^3 + z^3 = 3z + 3x + 4 \\
x^3 + z^3 = 3x + 3y + 4
\end{cases}
\]

1. **Assume \( x \ge y \ge z \):**
   - From the first equation: \( x^3 + y^3 = 3y + 3z + 4 \)
   - From the second equation: \( y^3 + z^3 = 3z + 3x + 4 \)
   - From the third equation: \( x^3 + z^3 = 3x + 3y + 4 \)

2. **Compare the equations:**
   - Since \( x \ge y \ge z \), we have:
     \[
     x^3 + z^3 \ge y^3 + z^3 \implies 3x + 3y + 4 \ge 3z + 3x + 4 \implies y \ge z
     \]
   - Similarly, from \( x^3 + y^3 \ge y^3 + z^3 \):
     \[
     3y + 3z + 4 \ge 3z + 3x + 4 \implies y \ge x
     \]

3. **Conclude \( x = y = z \):**
   - From the inequalities \( x \ge y \ge z \ge x \), we deduce \( x = y = z \).

4. **Substitute \( x = y = z \) into any equation:**
   - Using \( x = y = z \) in the first equation:
     \[
     x^3 + x^3 = 3x + 3x + 4 \implies 2x^3 = 6x + 4
     \]
   - Simplify the equation:
     \[
     2x^3 = 6x + 4 \implies x^3 = 3x + 2
     \]
   - Solve the cubic equation \( x^3 - 3x - 2 = 0 \):
     - Factorize \( x^3 - 3x - 2 = 0 \):
       \[
       (x - 2)(x^2 + 2x + 1) = 0 \implies (x - 2)(x + 1)^2 = 0
       \]
     - The solutions are \( x = 2 \) and \( x = -1 \).

5. **Verify the solutions:**
   - For \( x = 2 \):
     \[
     2^3 + 2^3 = 3(2) + 3(2) + 4 \implies 16 = 16 \quad \text{(True)}
     \]
   - For \( x = -1 \):
     \[
     (-1)^3 + (-1)^3 = 3(-1) + 3(-1) + 4 \implies -2 = -2 \quad \text{(True)}
     \]

Thus, the solutions to the system are \( x = y = z = 2 \) and \( x = y = z = -1 \).

The final answer is \( \boxed{ x = y = z = 2 } \) and \( x = y = z = -1 \).