$f(x)=x^3-ax^2-ax+1,\;\;|x_1|+|x_2|+|x_3|=3$ 
Find all $a\in \mathbb R$ such that the zero-points of the function
  $$f(x)=x^3-ax^2-ax+1$$ Satisfy the equation: $$|x_1|+|x_2|+|x_3|=3$$

My work:
$$
\begin{split}
f(x) &= x^3-ax^2-ax+1 \\
     &= x^3+1-ax(x+1)\\
     &= (x+1)(x^2-x+1-ax) \\
     &=(x+1)(x^2-(a+1)x+1) \\
x_1  &= -1
\end{split}
$$
By Viéta's formulae:
$$
\begin{split}
x_2 +     x_3 &= a+1 \\
x_2 \cdot x_3 &= 1\implies x_2\;\&\;x_3>0 \lor x_2\;\&\;x_3<0\\
a &= 1\implies x_2=x_3=-1 \leftarrow \Delta=0 \\
&\text{square of a binomial, double solution}
\end{split}
$$
$$$$
Then it really is:
for $x_1=x_2=x_3=-1$
$$|x_1|+|x_2|+|x_3|=3$$
Is this correct?

$ x_1 = -1 $, we have 
$$ |x_2| + |x_3| = 2 $$
now, if $ x_2, x_3 $ real Number
can be four solution.


*

*$ x_2 > 0, x_3 > 0 $
$x_2 + x_3 = a+1$
$a =1$


*

*$ x_2 < 0, x_3 < 0 $
$x_2 + x_3 = -(a+1)$
$a =-3$


*

*$ x_2 < 0, x_3 > 0 $ or $ x_2 > 0, x_3 < 0 $
$ x_2 - x_3 = +- \sqrt{(a+1)^2 - 4} $
If $x_2 = x_3$
So,
$ (a+1)^2 - 4 = 0 $
$ a = -3 or a = 1$
if not ? $ a < -3 or a > 1$
