Problem: Sum of absolute values of polynomial roots Can you please give me some hints as to how I might approach this problem? Thanks!

Given the polynomial $f = 2X^3 - aX^2 - aX + 2, a \mathbb \in R$ and roots $x_1, x_2$ and $x_3,$ find $a$ such that $|x_1| + |x_2| + |x_3| = 3.$

Edit: We know $-1$ is one of the roots of that polynomial, regardless of the value of $a$. So, in essence, what we have to demonstrate is that $|x_2| + |x_3| = 2.$
 A: HINT
$|x_1| + |x_2| + |x_3| \geq 3 \sqrt[3]{|x_1||x_2||x_3|} = 3$
Hence, we have $|x_1| + |x_2| + |x_3| \geq 3$. Equality holds implies $|x_1| = |x_2| = |x_3| = 1$
We have $x_1 + x_2 + x_3$, $x_1x_2 + x_2x_3 + x_3x_2$ and $x_1 x_2 x_3$ to be real,  and further $x=-1$ satisfies the equation.
Hence $f(x) = 2 \left( x+1 \right) \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$. We have $x_3 = -1$. And $|x_1| = 1$ and $|x_2| = 1$. Hence, $x_1 = e^{i \theta}$ and $x_2 = e^{i \phi}$.
$(x-x_1)(x-x_2) = \left( x^2-\left(1 + \frac{a}{2} \right)x + 1 \right)$
$x_1 x_2 = 1 \implies \phi = -\theta$
$x_1 + x_2 = 2 \cos(\theta) = 1 + \frac{a}{2}$
Hence, $\frac{a}{2} = 2 \cos(\theta) - 1 \implies a = 4 \cos(\theta) - 2$.
Hence, $a = 4 \cos(\theta) - 2$ and the roots are $-1,e^{i \theta},e^{-i \theta}$
A: We first have the following factorization,
$ 2x^3 - ax^2 - ax + 2 = (x+1)(2x^2 - (2+a)x + 2).$
Suppose $x_1 = -1$, then $|x_2| + |x_3| = 2$ and they are solutions to the quadratic equation $2x^2 - (2+a)x + 2 = 0$. Hence,
$ x_2 + x_3 = 1 + a/2 $ and 
$ x_2 x_3 = 1$.
By observing the second equation, $x_2$ and $x_3$ are either both positive or both negative, so we have
$1 + a/2 = x_2 + x_3 = 2$ or $-2$. 
Therefore, $a = 2$ or $-6$.
A: You can find such $a$ by inspection: Put $a=2$ and the roots are given by
$$x^3 - x^2 - x + 1 = 0 \Leftrightarrow (x-1)^2(x+1) = 0.$$
A: The way to solve this is actually quite simple, apparently:
$$|x_2| + |x_3| =2 \implies -2 \le x_1 + x_1 \le 2$$ 
