What is the Sum of Roots for $P(x)=x^3-2x^2-x+1$ with $x_1>x_2>x_3$?

[anemone]

Let $x_1,\,x_2,\,x_3$ be the three real roots of $P(x)=x^3-2x^2-x+1$ such that $x_1>x_2>x_3$.

Evaluate $x_1^2x_2+x_2^2x_3+x_3^2x_1$.

 

[anemone]

Subtle hint:

Spoiler

Make full use of the expressions $x_1^2x_2+x_2^2x_3+x_3^2x_1$ and $x_1x_2^2+x_2x_3^2+x_3x_1^2$.

 

[anemone]

Subtle hint:

Spoiler

Make full use of the expressions $x_1^2x_2+x_2^2x_3+x_3^2x_1$ and $x_1x_2^2+x_2x_3^2+x_3x_1^2$.

Follow-up hint:

Spoiler

Relate $x_1^2x_2+x_2^2x_3+x_3^2x_1$ and $x_1x_2^2+x_2x_3^2+x_3x_1^2$ with

1. $x_1+x_2+x_3$,

2. $x_1x_2+x_2x_3+x_3x_1$,

3. $x_1x_2x_3$.

 

[Greg]

Follow-up hint:

Spoiler

Relate $x_1^2x_2+x_2^2x_3+x_3^2x_1$ and $x_1x_2^2+x_2x_3^2+x_3x_1^2$ with

1. $x_1+x_2+x_3$,

2. $x_1x_2+x_2x_3+x_3x_1$,

3. $x_1x_2x_3$.

Spoiler

I've managed to get as far as

$x_1^2x_2+x_2^2x_3+x_3^2x_1+x_1x_2^2+x_2x_3^2+x_3x_1^2=1$

Currently, I don't see a way of progressing any further.

 

[kaliprasad]

Spoiler

Let $S_1=x_1^2x_2+x_2^2x_3+x_3^2x_1$
and $S_2 = x_1x_2^2 + x_2x_3^2 + x_3x_1^2$
(slight cheat here used a computer)
We have $P(-.9)=-.449,P(-.8)=.008,P(.5)= .125,P(.6)=-.104, P(2.2)=-.232, P(2.3) = .287$
It follows that the roots satisfy $-.9<x_3<-.8, .5<x_2<.6$ and $2.2<x_1<2.3$
This implies that $x_1^2x_2 + x_2^2x_3+ x_3^2x_1 > 0$
or $S_1 > 0$
Now $S_1 + S_2 = (x_1^2x_2+x_2^2x_3+x_3^2x_1) + (x_1x_2^2 + x_2x_3^2 + x_3x_1^2)$
$= x_1x_2(x_1+x_2) + x_2x_3(x_3+x_2)+ x_3x_1(x_3+x_1)$
$= (x_2+x_3+x_1)(x_1x_2+x_2x_3+ x_3x_1) - 3 x_1x_2x_3$
$= 2 * (-1) + 3 = 1$
The product $S_1S_2$ is
$S_1S_2 = (x_1^2x_2 + x_2^2x_3 + x_3^2x_1^2)(x_1x_2^2 + x_2x_3^2 + x_3x_1^2)$
= $x_1^3x_2^3 + x_2^3x_3^3 + x_3^3x_1^3 + 3x_1^2x_2^2 x_3^2 + x_1x_2x_3(x_1^3 + x_2^3 + x_3^3)$
Now we need to evaluate $x_1^3x_2^3 + x_2^3x_3^3 + x_3^3x_1^3$ and so on
To evaluate that, notice that the equation with roots $x_1^3,x_2^3,x_3^3$ by
letting $y = x^3 $in the original equation, which then becomes $y+1=2(\sqrt[3]y)^2 + \sqrt[3]y$
Cube both sides to get
$y^3+3y^2+ 3y + 1 = 8y^2 + y + 6y(y+1)$
or $y^3-11y^2-4y+1=0$
Therefore $x_1^3+x_2^3+x_3^3=11$ , $x_1^3x_2^3+x_2^3x_3^3 + x_3^3x_1^3 = -4$
Substitute those values into the above displayed equation to get
$S_1S_2=-4+3-11=-12$
Thus the equation with roots $S_1$ and $S_2$ are
$x^2 -x - 12 = 0$ or $(x-4)(x+3) = 0$ so x = 4 or -3
as $S_1 > 0$ so it is 4
or $x_1^2x_2 + x_2^2x_3+ x_3^2x_1 = 4$

 

[Albert1]

prove of $S_1>0$ without using a computer

Spoiler

$P(-1)=-1<0, P(0)=1>0, P(1)=-1<0,P(3)=7>0$
from Descartes' rule of signs
P(X) has two positive roots and one negative root
$\therefore x_1>1>x_2>0>x_3$
it is clear that:$x_1^2x_2>0,x_2^2x_3<0,$ and $x_3^2x_1>0$
this implies $S_1=x_1^2x_2+x_2^2x_3+x_3^2x_1>0$
for $x_1^2x_2>\left | x_2^2x_3 \right |$

 

[anemone]

Hi all!

I've been very busy in the past few days and I am sick today...so sorry all for the late reply!

Thanks kaliprasad for your neat and elegant solution and thanks to greg1313 and Albert as well for your participation!

Spoiler

Yes, greg1313, if we let $S_1=x_1^2x_2+x_2^2x_3+x_3^2x_1$ and $S_2=x_1x_2^2+x_2x_3^2+x_3x_1^2$, then $S_1+S_2=1$, but note that we could also draw some useful information if we subtract $S_2$ from $S_1$ as $S_1-S_2=(x_1-x_2)(x_2-x_3)(x_1-x_3)>0\implies S_1>S_2$ and we then see if we can write the product of $S_1$ and $S_2$ using only the expressions below:

1. $x_1+x_2+x_3$,

2. $x_1x_2+x_2x_3+x_3x_1$,

3. $x_1x_2x_3$.

$\begin{align*}S_1S_2&=(x_1^2x_2 + x_2^2x_3 + x_3^2x_1^2)(x_1x_2^2 + x_2x_3^2 + x_3x_1^2)\\&=x_1^3x_2^3 + x_2^3x_3^3 + x_3^3x_1^3 + 3x_1^2x_2^2 x_3^2 + x_1x_2x_3(x_1^3 + x_2^3 + x_3^3)\\&=(x_1x_2+x_2x_3+x_3x_1)^3+(x_1+x_2+x_3)^3(x_1x_2x_3)+9(x_1x_2x_3)^2-6(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)(x_1x_2x_3)\\&=-12\end{align*}$

Therefore, we get $S_1=4,\,S_2=-3$ and we're hence done! :D

 

[kaliprasad]

Hi all!

I've been very busy in the past few days and I am sick today...so sorry all for the late reply!

wish you a speedy recovery

 

[anemone]

wish you a speedy recovery

Thank you, Kali!