Find all real numbers with the following property: the difference of its cube and
its square is equal to the square of the difference of its square and the number itself.
1. Let the number be \( x \). The problem states that the difference of its cube and its square is equal to the square of the difference of its square and the number itself. This can be written as:
   \[
   x^3 - x^2 = (x^2 - x)^2
   \]

2. Simplify the right-hand side:
   \[
   (x^2 - x)^2 = x^4 - 2x^3 + x^2
   \]

3. Substitute this back into the equation:
   \[
   x^3 - x^2 = x^4 - 2x^3 + x^2
   \]

4. Rearrange all terms to one side of the equation:
   \[
   x^4 - 3x^3 + 2x^2 = 0
   \]

5. Factor out the common term \( x^2 \):
   \[
   x^2 (x^2 - 3x + 2) = 0
   \]

6. This gives us two factors to solve:
   \[
   x^2 = 0 \quad \text{or} \quad x^2 - 3x + 2 = 0
   \]

7. Solve the first factor:
   \[
   x^2 = 0 \implies x = 0
   \]

8. Solve the quadratic equation \( x^2 - 3x + 2 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{3 \pm \sqrt{9 - 8}}{2} = \frac{3 \pm 1}{2}
   \]
   \[
   x = \frac{4}{2} = 2 \quad \text{or} \quad x = \frac{2}{2} = 1
   \]

9. Therefore, the solutions are:
   \[
   x = 0, 1, 2
   \]

The final answer is \(\boxed{0, 1, 2}\)