Let's analyze the equation $X! = x(x^2 - 1)$.

• $X!$ represents the factorial of $X$, which means $X! = X \times (X-1) \times (X-2) \times \dots \times 1$.
• The right-hand side of the equation is a polynomial expression $x(x^2 - 1)$, which simplifies to $x^3 - x$.

So, we need to find values of $X$ and $x$ that satisfy the equation:

$X! = x(x^2 - 1)$

This implies:

$X! = x^3 - x$

At this point, solving for specific values of $X$ and $x$ depends on whether $X$ is a fixed number or if we are solving for both variables. Would you like to explore potential solutions for specific values of $X$, or do you need help solving it algebraically for general $X$?