Find all real numbers $x, y, z$ that satisfy the following system
$$\sqrt{x^3 - y} = z - 1$$
$$\sqrt{y^3 - z} = x - 1$$
$$\sqrt{z^3 - x} = y - 1$$
To solve the system of equations:
\[
\sqrt{x^3 - y} = z - 1
\]
\[
\sqrt{y^3 - z} = x - 1
\]
\[
\sqrt{z^3 - x} = y - 1
\]

we will proceed step-by-step to find all real numbers \(x, y, z\) that satisfy these equations.

1. **Assume \(x = y = z\):**
   Let's start by assuming \(x = y = z\). This simplifies the system to:
   \[
   \sqrt{x^3 - x} = x - 1
   \]
   \[
   \sqrt{x^3 - x} = x - 1
   \]
   \[
   \sqrt{x^3 - x} = x - 1
   \]
   Since all three equations are identical under this assumption, we only need to solve one of them:
   \[
   \sqrt{x^3 - x} = x - 1
   \]

2. **Square both sides:**
   To eliminate the square root, we square both sides of the equation:
   \[
   (\sqrt{x^3 - x})^2 = (x - 1)^2
   \]
   \[
   x^3 - x = x^2 - 2x + 1
   \]

3. **Rearrange the equation:**
   Bring all terms to one side to form a polynomial equation:
   \[
   x^3 - x - x^2 + 2x - 1 = 0
   \]
   \[
   x^3 - x^2 + x - 1 = 0
   \]

4. **Factor the polynomial:**
   Notice that \(x = 1\) is a root of the polynomial:
   \[
   1^3 - 1^2 + 1 - 1 = 0
   \]
   Therefore, \(x - 1\) is a factor of the polynomial. We can factor the polynomial as:
   \[
   x^3 - x^2 + x - 1 = (x - 1)(x^2 + 1)
   \]

5. **Solve for the roots:**
   Set each factor equal to zero:
   \[
   x - 1 = 0 \quad \text{or} \quad x^2 + 1 = 0
   \]
   \[
   x = 1 \quad \text{or} \quad x^2 = -1
   \]
   Since \(x^2 = -1\) has no real solutions, the only real solution is:
   \[
   x = 1
   \]

6. **Verify the solution:**
   If \(x = 1\), then \(y = 1\) and \(z = 1\). Substitute \(x = y = z = 1\) back into the original equations:
   \[
   \sqrt{1^3 - 1} = 1 - 1 \quad \Rightarrow \quad \sqrt{0} = 0 \quad \Rightarrow \quad 0 = 0
   \]
   \[
   \sqrt{1^3 - 1} = 1 - 1 \quad \Rightarrow \quad \sqrt{0} = 0 \quad \Rightarrow \quad 0 = 0
   \]
   \[
   \sqrt{1^3 - 1} = 1 - 1 \quad \Rightarrow \quad \sqrt{0} = 0 \quad \Rightarrow \quad 0 = 0
   \]
   All three equations are satisfied.

Therefore, the only solution is \((x, y, z) = (1, 1, 1)\).

The final answer is \( \boxed{ (1, 1, 1) } \)