Solvablility of a Galois group Question: For a positive integer $k$, consider the polynomial $f(x)=x^6+3kx^4+3x^3+3kx^2+1\in\mathbb{Q}[x]$. Suppose that $F$ is a splitting field of $f(x)$, show that the Galois group $\mathrm{Gal}(F/\mathbb{Q})$ is solvable.
Comments: It is not hard to check that $f(x)$ is irreducible in $\mathbb{Q}[x]$, so $\mathrm{Gal}(F/\mathbb{Q})=S_6$ or $\mathrm{Gal}(F/\mathbb{Q})\subseteq A_6$. Then how to show that it is a solvable subgroup in $A_6$?
 A: Your polynomial is palindromic (the sequence of its coefficients does not change, if you revert it). More precisely,
$$
f(x)=x^6f(\frac1x).
$$
This implies that if $\alpha$ is a zero of $f(x)$, so is $1/\alpha$.
A common trick when dealing with palindromic polynomials is to introduce the variable
$$
u=x+\frac1x.
$$
We can rewrite
$$
\begin{aligned}
\frac1{x^3}f(x)&=x^3+\frac1{x^3}+3k(x+\frac1x)+3\\
&=(u^3-3u)+3ku+1\\
&=u^3+3(k-1)u+3
\end{aligned}
$$
using the new variable $u$.
The polynomial $g(u)=u^3+3(k-1)u+3$ is a cubic. It is also irreducible by Eisenstein ($p=3$). Anyway, its splitting field $K$ is a solvable extension of $\Bbb{Q}$ of degree $3$ or $6$. Clearly $K\subseteq F$, because the zeros of $g(u)$ have the form $u=\alpha+1/\alpha$, where $\alpha$ is a zero of $f(x)$. Furthermore, 
if the zeros of $g(u)$ are $u_1,u_2,u_3$, we get $F$ by adjoining the zeros of the quadratic polynomials
$$
h_i(x)=x^2-u_ix+1\in K[x],
$$
$i=1,2,3$ to the field $K$. We also have the factorization
$$
f(x)=h_1(x)h_2(x)h_3(x).
$$
This means that $[F:K]=2^t$, where $t=1,2$ or $3$. We cannot easily tell the value of $t$, because adjoining the zeros of some of $h_i$, may bring the rest of them as well. However, we can now say that $[F:\Bbb{Q}]=2^a3$ with $a\in\{0,1,2,3,4\}$. Also (thanks Mercio!) the solvability of $F/\Bbb{Q}$ follows from the solvability of $K/\Bbb{Q}$ and the fact that we get $F$ by adjoining roots of quadratics to $K$.
