Seeking an explanation for a peculiar factorization 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!
 A: You have reciprocal polynomial and after standart 
change of variable $t=x+1/x$ you'll get the cubic equation
$$t^3-3t^2+(c-3)t+11-2c=0.$$
A: Ah, I have used something like this as a math problem in a math competition!
Note that you have inversion symmetry in your polynomial, meaning that $p(x)$ and $x^6p(1/x)$ have the same roots. This means that if $x$ is a root, then so is $1/x$. You can sort of "fold" these roots together, which is the reason for the cubic. 
Also, you have the classical symmetry, that roots come in conjugate pairs. This adds a lot of restriction to your polynomial: for example, if it does not have any positive real roots, it has to have a root on the unit circle.
Addition:
To expand on the fold comment, we map the unit circle to the real line, using the Möbius map $x \to (t-1)/(t+1)$. This gives
$$
\frac{t^6 + (4 c-15) t^4 + (75- 8 c)t^2 + 4c+3}{(t+1)^6}
$$
which we are interested in the roots of. Note that the numerator is even, which is the fold I was mentioning. Substituting $t^2 = s$ in the numerator gives
$$s^3 + (4c-15) s^2 + (75-8 c) s + 4c + 3$$ 
I don't know how related this cubic is to yours - but this shows that a connection via a cubic is to be expected for a palindromic sextic.
Addition II: One can recover your sextic from a Gröbner basis computation simply by the following Mathematica code:
GroebnerBasis[
 {a^3 + (3 - c) a^2 + 3 a - 1,
  x^2 - x + a}, {a, c}, {a}]

Basically, this verifies that the roots of your sextic is given by taking a root from a cubic, and plugging in as a coefficient in a quadratic. 
The polynomial 
$$
1 + b x + c x^2 + (-1 - 2 b - 2 c) x^3 + (3 + b + c) x^4 - 3 x^5 + x^6
$$
has similar properties as the second one you give, and I produced it via
GroebnerBasis[{a^3 + (-b - c) a^2 - b a - 1,x^2 -  x + a }, {b, c}, a}];

Note that the quadratic forces the total sum of the roots to be equal to $1$, since the coefficient of $x$ is $-1$.
Finally, note that the roots of $x^2-x+(a\pm i b)=0$ gives $4$ roots that appear in similar configuration as in your second question.
