The equation you provided is:

$x! = x^3 - x$

This is a non-standard equation involving a factorial expression $x!$ and a polynomial $x^3 - x$.

To solve this equation, let's explore possible integer values for $x$, as factorials are only defined for non-negative integers.

Step-by-step approach:

1. Check small integer values for $x$ since the factorial function grows rapidly:

   • For $x = 0$: $0! = 1 \quad \text{and} \quad 0^3 - 0 = 0$ So, $1 \neq 0$. Hence, $x = 0$ is not a solution.

   • For $x = 1$: $1! = 1 \quad \text{and} \quad 1^3 - 1 = 0$ So, $1 \neq 0$. Hence, $x = 1$ is not a solution.

   • For $x = 2$: $2! = 2 \quad \text{and} \quad 2^3 - 2 = 6$ So, $2 \neq 6$. Hence, $x = 2$ is not a solution.

   • For $x = 3$: $3! = 6 \quad \text{and} \quad 3^3 - 3 = 24$ So, $6 \neq 24$. Hence, $x = 3$ is not a solution.

   • For $x = 4$: $4! = 24 \quad \text{and} \quad 4^3 - 4 = 60$ So, $24 \neq 60$. Hence, $x = 4$ is not a solution.

   • For $x = 5$: $5! = 120 \quad \text{and} \quad 5^3 - 5 = 120$ So, $120 = 120$. Hence, $x = 5$ is a solution.

Conclusion:

The only solution to the equation $x! = x^3 - x$ is $x = 5$.

Would you like to dive deeper into factorial properties, explore other types of equations, or have any specific questions about this problem? Here are some related questions to further explore:

1. How do factorials grow compared to polynomials?
2. What would happen if the equation involved a higher power of $x$?
3. How would the equation change if we used negative factorials?
4. What are some common techniques for solving equations involving factorials?
5. How can this equation be generalized for other operations or constraints?

Tip: Factorial functions grow very fast, which is why when solving equations like this, we often start by testing small integer values to find potential solutions.