TITLE: Solve this equation: $3^{3x} - 3^x = (3x)!$
QUESTION [17 upvotes]: 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.

REPLY [4 votes]: 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 $x=2.$