$f(x)=x^3-ax^2-ax+1,\;\;|x_1|+|x_2|+|x_3|=3$

Question

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?

Answer 1

$ 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$