TITLE: Seeking an explanation for a peculiar factorization
QUESTION [6 upvotes]: Recently during my work, I encountered the following family of sextic polynomials
$$\displaystyle x^6 - 3x^5 + cx^4 + (5 - 2c)x^3 + cx^2 - 3x + 1.$$
By plugging in various values of $c$, I noticed that this sextic always has all real roots or all non-real roots. When the roots are all complex, I noticed the following: there is always a conjugate pair $\alpha, \overline{\alpha}$ such that the real part of $\alpha$ is $1/2$, the other two pairs $\beta, \overline{\beta}$ and $\gamma, \overline{\gamma}$ have the same imaginary parts, and the real part of $\beta$ and $\gamma$ always sum to $1$. Using this data, I formulated relations between the roots and found that in fact the sextic always factors as follows:
 $$\displaystyle x^6 - 3x^5 + cx^4 + (5 - 2c)x^3 + cx^2 - 3x + 1 =$$
$$\displaystyle (x^2 - x + a)(x^2 - x/a + 1/a)(x^2 - (2 - 1/a)x + 1),$$
where $a$ is a real root of the cubic equation
$$\displaystyle x^3 + (3 - c)x^2 + 3x - 1 = 0.$$
My proof is by brute force. Is there any more enlightened explanation of this fact?
Edit: in my latest experiments, I have discovered that polynomials of the shape
$$\displaystyle x^6 + bx^5 + cx^4 + (2c + 10 - 5b)x^3 + (c + 15 - 5b)x^2 + (6 - b)x + 1$$ 
also have the strange property that it either has all real roots or all complex roots, and if all of the roots are complex, then there is a pair with real part equal to $1/2$ and the other two pairs have real parts summing to one and identical imaginary parts. This should lead to a similar factorization once one works out the details, but there is no obvious folding that occurs!

REPLY [6 votes]: You have reciprocal polynomial and after standart 
change of variable $t=x+1/x$ you'll get the cubic equation
$$t^3-3t^2+(c-3)t+11-2c=0.$$