The curve discussed in this OP's post,
is birationally equivalent to an elliptic curve. Following E. Delanoy's post, let $G$ be the set of rational numbers that solve $(1)$. Courtesy of Aretino's comment, it is known that $G$ is invariant by the transformation,
thus both $a$ and $f(a)$ are solutions to $(1)$. Equivalently,
Q: Let $P(a)$ be the irreducible nonic denominator. Why does it have a solvable Galois group?
In other words, the eqn $P(a) = 0$ is solvable in radicals. There is an online Magma calculator that computes the Galois group and the command is,
Z := Integers(); P < x > := PolynomialRing(Z); f := -216 + 1296*x^2 - 2160*x^3 + 1296*x^4 - 108*x^5 - 234*x^6 + 108*x^7 - 18*x^8 + x^9; G, R := GaloisGroup(f); G;
It says that the order is $54$ and hence is solvable.
P.S. This is the second time within a short period that I've come across a polynomial identity that surprisingly has a solvable Galois group. (The explicit identity is given in the first link above.) See also the recent post, Why does $x^2+47y^2 = z^5$ involve solvable quintics?