Find all real numbers $x,y,z$ such that satisfied the following equalities at same time:

$\sqrt{x^3-y}=z-1 \wedge \sqrt{y^3-z}=x-1\wedge \sqrt{z^3-x}=y-1$
1. Given the system of equations:
   \[
   \sqrt{x^3 - y} = z - 1 \quad \text{and} \quad \sqrt{y^3 - z} = x - 1 \quad \text{and} \quad \sqrt{z^3 - x} = y - 1
   \]
   We start by noting that the square root function implies non-negativity:
   \[
   \sqrt{x^3 - y} \geq 0 \implies x^3 - y \geq 0 \implies y \leq x^3
   \]
   Similarly,
   \[
   \sqrt{y^3 - z} \geq 0 \implies y^3 - z \geq 0 \implies z \leq y^3
   \]
   and
   \[
   \sqrt{z^3 - x} \geq 0 \implies z^3 - x \geq 0 \implies x \leq z^3
   \]

2. From the non-negativity of the square roots, we also have:
   \[
   \sqrt{x^3 - y} = z - 1 \implies z \geq 1
   \]
   \[
   \sqrt{y^3 - z} = x - 1 \implies x \geq 1
   \]
   \[
   \sqrt{z^3 - x} = y - 1 \implies y \geq 1
   \]
   Therefore, we have \(x, y, z \geq 1\).

3. Next, we rewrite the original equations in terms of \(x, y, z\):
   \[
   x^3 - y = (z - 1)^2 \implies x^3 - y = z^2 - 2z + 1
   \]
   \[
   y^3 - z = (x - 1)^2 \implies y^3 - z = x^2 - 2x + 1
   \]
   \[
   z^3 - x = (y - 1)^2 \implies z^3 - x = y^2 - 2y + 1
   \]

4. Adding these three equations:
   \[
   (x^3 - y) + (y^3 - z) + (z^3 - x) = (z^2 - 2z + 1) + (x^2 - 2x + 1) + (y^2 - 2y + 1)
   \]
   Simplifying, we get:
   \[
   x^3 + y^3 + z^3 - (x + y + z) = x^2 + y^2 + z^2 - 2(x + y + z) + 3
   \]
   \[
   x^3 + y^3 + z^3 - x - y - z = x^2 + y^2 + z^2 - 2x - 2y - 2z + 3
   \]
   \[
   x^3 + y^3 + z^3 - x - y - z = x^2 + y^2 + z^2 - x - y - z - (x + y + z) + 3
   \]
   \[
   x^3 + y^3 + z^3 - x - y - z = x^2 + y^2 + z^2 - 3(x + y + z) + 3
   \]

5. Rearranging terms, we get:
   \[
   x^3 + y^3 + z^3 - x^2 - y^2 - z^2 + 2(x + y + z) = 3
   \]
   We can rewrite this as:
   \[
   (x^3 - x^2 + x) + (y^3 - y^2 + y) + (z^3 - z^2 + z) = 3
   \]

6. Notice that for \(x \geq 1\), \(y \geq 1\), and \(z \geq 1\), each term \(x^3 - x^2 + x \geq 1\), \(y^3 - y^2 + y \geq 1\), and \(z^3 - z^2 + z \geq 1\). Therefore:
   \[
   (x^3 - x^2 + x) + (y^3 - y^2 + y) + (z^3 - z^2 + z) \geq 3
   \]
   Since we have equality, it must be that:
   \[
   x^3 - x^2 + x = 1, \quad y^3 - y^2 + y = 1, \quad z^3 - z^2 + z = 1
   \]

7. Solving \(x^3 - x^2 + x = 1\) for \(x \geq 1\), we find that \(x = 1\). Similarly, \(y = 1\) and \(z = 1\).

Conclusion:
\[
\boxed{x = y = z = 1}
\]