Solving $a+b+c = abc = 6$ in the rationals entails solving,

$$-24a+36a^2-12a^3+a^4=z^2\tag1$$

which is birationally equivalent to an *elliptic curve*. It can be shown that if $a$ is a solution, then so is the transformation,

$$f(a)=\frac{-54a(-6+12a-6a^2+a^3)^2}
{−216+1296a^2−2160a^3+1296a^4−108a^5−234a^6+108a^7−18a^8+a^9}$$

The irreducible nonic has a solvable Galois group with order $54$. (A related discussion is in this [MSE post][1].) More generally, if $a+b+c = abc = n$ so,

$$-4na+n^2a^2-2na^3+a^4=z^2\tag2$$

Define $u=\large{\frac{-a^3+na^2+n}{a}}$. If $a$ is a solution to $(2)$, then so is,

$$f(a) = \frac{n(n^2-3u)^2}{u^3-9n^2u+n^2(n^2+27)}$$

Expanding the denominator, one gets a nonic in $a$ with a solvable Galois group. (The particular example was just the case $n=6$.) Even more generally, given ***any elliptic curve***,

$$F(x) = y^2\tag3$$

and the transformation,

$$f(x) = \frac{g(x)}{h(x)}$$

>**Question 1:** If both $x$ and $f(x)$ are rational solutions to $(3)$, does that imply $g(x)$ and $h(x)$ have solvable Galois groups?

*In response to a comment:*

>**Question 2:** Define $(3)$ as an elliptic curve with an infinite number of distinct rational points. Assume the set $S$ of transformations $f_i(x) = \frac{g_i(x)}{h_i(x)}$ such that both $x$ and $f_i(x)$ are rational solutions to $(3)$. In this set $S$, can there be always *at least one* non-trivial transformation such that its $g_i(x)$ and $h_i(x)$ have solvable Galois groups?

  [1]: http://math.stackexchange.com/questions/1391806/