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 $x_1^2x_2 + x_2^2x_3+ x_3^2x_1 = 4$