For each integer $n \ge 3$ solve in real numbers the system of equations:
$$\begin{cases}  x_1^3 = x_2 + x_3 + 1 \\...\\x_{n-1}^3 = x_n+ x_1 + 1\\x_{n}^3 = x_1+ x_2 + 1 \end{cases}$$
1. **Identify the system of equations:**
   \[
   \begin{cases}
   x_1^3 = x_2 + x_3 + 1 \\
   x_2^3 = x_3 + x_4 + 1 \\
   \vdots \\
   x_{n-1}^3 = x_n + x_1 + 1 \\
   x_n^3 = x_1 + x_2 + 1
   \end{cases}
   \]

2. **Assume a solution form:**
   Let \( t \) be any root of the equation \( t^3 = 2t + 1 \). We claim that \( x_i = t \) for all \( 1 \le i \le n \) is the only solution.

3. **Find the roots of the polynomial:**
   Consider the polynomial equation:
   \[
   t^3 = 2t + 1 \implies t^3 - 2t - 1 = 0
   \]
   Factorize the polynomial:
   \[
   t^3 - 2t - 1 = (t + 1)(t^2 - t - 1) = 0
   \]
   The roots are:
   \[
   t = -1 \quad \text{or} \quad t^2 - t - 1 = 0
   \]
   Solving the quadratic equation \( t^2 - t - 1 = 0 \):
   \[
   t = \frac{1 \pm \sqrt{5}}{2}
   \]
   Therefore, the roots are:
   \[
   t = -1, \quad t = \frac{1 + \sqrt{5}}{2}, \quad t = \frac{1 - \sqrt{5}}{2}
   \]

4. **Verify the solution:**
   Assume \( x_i = t \) for all \( 1 \le i \le n \). Substitute \( x_i = t \) into the system of equations:
   \[
   t^3 = t + t + 1 = 2t + 1
   \]
   This is satisfied by the roots \( t = -1 \), \( t = \frac{1 + \sqrt{5}}{2} \), and \( t = \frac{1 - \sqrt{5}}{2} \).

5. **Uniqueness of the solution:**
   Let \( x_i = \max_{j \le n} x_j \). Using \( x_i^3 \ge x_{i-1}^3 \), we find \( x_{i+2} = x_i \). Repeating this, we find \( x_i = x_{i+2} = x_{i+4} = \cdots \). Now, \( x_{i-1}^3 = x_i + x_{i+1} + 1 = x_{i+2} + x_i + 1 = x_i^3 \) yields, together with the monotonicity of \( t \mapsto t^3 \), \( x_{i-1} = x_i \). Iterating the idea above, we find all \( x_i \) are in fact equal. Setting \( x_i = t \), we arrive at \( t^3 = 2t + 1 \) yielding \( (t + 1)(t^2 - t - 1) = 0 \).

The final answer is \( \boxed{ x_i = -1 } \) or \( x_i = \frac{1 + \sqrt{5}}{2} \) or \( x_i = \frac{1 - \sqrt{5}}{2} \) for all \( 1 \le i \le n \).