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Find all positive integers $x$ satisfying $x!=x^{3}-11 x+4$.
Intuitively, the factorial grows faster than the cube, so quite quickly $x^{3}-11 x+40$. Therefore, for $x \geq 5$, we have $x! > x^{3}-11 x+4$, so we cannot have equality. We need to test the values $0 \leq x \leq 4$:

| $x$ | $x!$ | $x^{3}-11 x+4$ | Equality? |
| :---: | :---: | :---: | :---: |
| 0 | 1 | 4 | no |
| 1 | 1 | -6 | no |
| 2 | 2 | -10 | no |
| 3 | 6 | -2 | no |
| 4 | 24 | 24 | YES |

Finally, there is a unique solution $x=4$.