Why does the elliptic curve for $a+b+c = abc = 6$ involve a solvable nonic? The curve discussed in this OP's post,
$$\color{brown}{-24a+36a^2-12a^3+a^4}=z^2\tag1$$
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,
$$f(a)=\frac{-54a(-6+12a-6a^2+a^3)^2}
{−216+1296a^2−2160a^3+1296a^4−108a^5−234a^6+108a^7−18a^8+a^9}$$
thus both $a$ and $f(a)$ are solutions to $(1)$. Equivalently,
$$f(a)= \frac{54a^2(-6+12a-6a^2+a^3)^2}{9a^2(-6+9a-6a^2+a^3)^2-(3-3a^2+a^3)^2\color{brown}{(-24a+36a^2-12a^3+a^4)}}$$

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? 
 A: Let $\infty_+$ and $\infty_-$ be the (real) points at infinity of the curve (corresponding to $y/x^2 = +1$ and $y/x^2 = -1$)
In general (unless $p^2=1$), a parabola $y = px^2+qx+r$ intersects the elliptic curve at $4$ points. More precisely, $Div(y-px^2-qx-r) = [P_1]+[P_2]+[P_3]+[P_4]-2[\infty_+]-2[\infty_-]$.
From a point $P$, you find the parabola that intersects the curve at $P$ three times and let $f(P)$ be the other intersection point. Which means that $f(P)$ is the point such that $3[P] + [f(P)] \sim 2[\infty_+] + 2[\infty_-]$ .
The $x$-coordinate of $f(P)$ only depends on the one of $P$ so this induces the rational function you have given, and you are asking about the points $P$ such that $3[P] \sim [\infty_+] + 2[\infty_-]$ (or $[\infty_+] + 2[\infty_-]$ but this will only switch the sigh of the $y$ coordinates).
Since $[\infty_+] + 2[\infty_-]$ is rational, the automorphism group of the $9$ points solution has to be included in the automorphism group of the affine plane over $\Bbb F_3$, i.e. $G = \left\{ \begin{pmatrix} a & b & c \\ d & e & f \\ 0 & 0 & 1 \end{pmatrix} \in GL_3(\Bbb F_3) \right\}$ which has order $432$ and turns out to be solvable.

In your case, the Galois group is coincidentally smaller : a choice of a point $P$ determines more than one point (it determines $3$ of them), so it should determine an automorphism of order $3$ of the curve. If you pick the point $(0,0)$ as the origin you can turn the elliptic curve into a group. 
In fact, since $f(\infty_+) = f(\infty_-) = (0,0)$, the points at infinity have order $3$, and so translation by $[\infty_+] - [0,0]$ is an order $3$ rational automorphism of the curve, which shows that we can replace $G$ with a subgroup $H$ of order $54$ (and every equation $f(a) = k$ for $k \in \Bbb Q \cup \{\infty\}$ has its galois group a subgroup of $H$).
I guess I'm blind and I only just now realised that $(a,b,c) \mapsto (b,c,a)$ is such an automorphism. So this should still work if you replace $6$ with $14$ or anything else.
A: Тhat is a beautiful solvable nonic! Here is its smallest real root:
With[{α = Sin[ArcSin[(1 + 3 × 3^(2/3))/8] / 3], 
      β = 3 + 9 × 3^(1/3) + 7 × 3^(2/3), 
      γ = 18 + 9 × 3^(1/3) + 2 × 3^(2/3)},
 With[{
   ρ = -51 - 33 × 3^(1/3) + 121 × 3^(2/3) - 8 α β + 4 α^2 (β - 60 × 3^(2/3)),
   σ = 12 - 9 × 3^(1/3) - 7 × 3^(2/3) + α γ + 2 α^2 β,
   τ = 53 + 24 × 3^(1/3) + 27 × 3^(2/3) - 8 α (11 + 3 × 3^(1/3) - 3^(2/3)) - 
           16 α^2 (1 + 3 × 3^(1/3) + 4 × 3^(2/3))},
        (90 (45 - 4 β) + 180 α (2 γ - 15) + 720 α^2 β - 
           5^(2/3) (135 × 6^(1/3) ρ^(1/3) + 36 σ τ^(1/3))) / 2025]]

