Find all polynomials $f$ that satisfy the equation $\frac{f(9 x)}{f(3 x)}=\frac{243 x-729}{x-81}$ for infinitely many values of $x$.
We have

$$
(x-81) f(9 x)=(243 x-729) f(3 x)
$$

for infinitely many values of $x$. Since both sides of this equation are polynomials, they must then be equal for all $x$.

Plugging in $x=3$, we get $f(27)=0$. Plugging in $x=9$ then gives $f(81)=0$. Plugging in $x=27$ then gives $f(243)=0$. Thus we have that $f(x)=g(x) \cdot(x-$ $27)(x-81)(x-243)$ for some polynomial $g(x)$.

Plugging this into the original equation, we get that for infinitely many $x$,

$243 \frac{x-3}{x-81}=\frac{f(9 x)}{f(3 x)}=\frac{g(9 x)}{g(3 x)} \cdot \frac{(9 x-27)(9 x-81)(9 x-243)}{(3 x-27)(3 x-81)(3 x-243)}=\frac{g(9 x)}{g(3 x)} \cdot 27 \frac{x-3}{x-81}$,
so $\frac{g(9 x)}{g(3 x)}=\frac{243}{27}=9$ and thus $f$ works if and only if $g(9 x)=9 g(3 x)$ for all $x$. Thus, the roots of $g(9 x)$ are those of $g(3 x)$, so they must all be zero, so $g$ must be of the form $a x^{n}$. Plugging in $x=1$ then gives $9^{n}=9 \cdot 3^{n}$, so $n=2$. Indeed, $g(x)=x^{2}$ works.

The solutions are then

$$
f(x)=a x^{2}(x-27)(x-81)(x-243)
$$

for real constants $a$.