Question

Let $x_1,x_2,x_{3}be3$ roots of the cubic $x^3-x-1=0$ then the expression $x_1{\left(x_2-x_3\right)}^2+x_2{\left(x_3-x_1\right)}^2+x_3{\left(x_1-x_2\right)}^2$ equal a rational number. Find the absolute value of the number.

Answer 1

Given equation.

$x^3-x-1=0$

If $x_1,x_2,x_3$ are roots then

$x_1+x_2+x_3=\frac{-constantterm}{coefficientof{x}^3}=1$

$x_1+x_2+x_3=\frac{-coefficientof{x}^2}{coefficientof{x}^3}=0$

$x_1{x}_2+x_2{x}_3+x_3{x}_1=\frac{coefficientofx}{coefficientof{x}^3}=-1$

divide last equation by $x_1,x_2,x_3$

$\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}=-1$

Expand the given equation

$x_1(x_2-x_3{)}^2+x_2(x_3-x_1{)}^2+x_3(x_2-x_1{)}^2=x_1{}^2{x}_2+x_1{}^2{x}_3+x_2{}^2{x}_1+x_2{}^2{x}_3+x_3{}^2{x}_1+x_3{}^2{x}_2-6x_1{x}_2{x}_3$

$x_1{x}_2{x}_3(\frac{x_1}{x_3}+\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_2}{x_1}+\frac{x_3}{x_2}+\frac{x_3}{x_1}-6)$

$=1\left(x_1\right(\frac{1}{x_2}+\frac{1}{x_3})+x_2(\frac{1}{x_3}+\frac{1}{x_1})+x_3(\frac{1}{x_2}+\frac{1}{x_1}-6)$

$=x_1(-1\frac{-1}{x_1})+x_2(-1\frac{-x}{x_2})+x_3(-1-\frac{-1}{x_3})-6$

$=-(x_1+x_2+x_3)-1-1-1-6=0-3-9$

$\therefore$ absolute value $|-9|=9$