Solve this equation: $3^{3x} - 3^x = (3x)!$ I have this equation:
$$3^{3x} - 3^x = (3x)!$$
We have to solve for $x$ integer. I did try to attempt but to no avail. I can't manipulate any side of this equation. I took common $3^x$ in the LHS of the equation and got a product: $(3^x) (3^{2x}-1)$ but I have no idea what to do in the RHS of the equation (which is a factorial). It looks like the answer is $x=2$ but I want to solve it algebraically.
Any hints/solution would be greatly appreciated.
 A: A factorial can be written as a product of three subsequent integers only in very few cases, in particular:
$$\begin{aligned}1\times 2\times 3&=3!\\2\times 3 \times 4=24&=4!\\4\times 5\times 6=120&=5! \end{aligned}$$ or
$$8\times 9\times 10=720=6! $$
LHS of the given equation is $$3^{3x} - 3^x = (3^x-1)(3^x)(3^x+1),$$ which is a product of subsequent integers where the mid-term is a power of $3.$ Comparing with the listed possibilities, only the last one satisfies. Here $3^x=9$ or $x=2.$
A: The main idea is that $n!>a^n$ for $n$ sufficiently large, so there is only a finite number of values to check.
In this problem, a simple mathematical induction shows that $n!>3^n$ for every $n\ge 7$. Therefore, for $x\ge 3$, we have $(3x)! > 3^{3x} > 3^{3x} - 3^x$.
Now, the equality is satisfied for $x=2$ ($3^6-3^2=720=6!$) but not for $0$ ($3^0-3^0=0\neq 0!$) or $1$ ($3^3-3^1=24\neq 3!$)
A: Here is a way by comparing exponents of $3$ on both sides for $x \geq 3$.
Notice that RHS is divisible by
$$
(3x)(3x-3)(3x-6)\cdots(3)=3^x(x(x-1)\cdots 1)=3^xx!.
$$
Because $x \geq 3$, this means $3 \mid x!$ and in turn $3^{x+1}$ divides RHS. But that means it must divide LHS as well, i.e. $3^{x+1} \mid  3^{3x} - 3^x = 3^x(3^{2x} - 1)$ (the equality you have found). But this means $3 \mid 3^{2x}-1$, impossible.
