$a^2b+b^2c+c^2a=3$ $?$

It is not correct. We have $a+b+c=2,ab+bc+ca=-1,abc=-1,c<0<b<a.$ Let $u=a^2b+b^c+c^2a>0,v=ab^2+bc^2+ca^2.$ u and v invariant by $s=(a\to b\to c),s^2$ But simmetric functions $u+v,uv$ invariant for all permutations. We have $u+v=(a+b+c)(ab+bc+ca)-3abc=1,$ and $uv=a^3b^3+b^3c^3+a^3c^3+3a^2b^2c^2+abc(a^3+b^3+c^3)$. $a^3b^3+b^3c^3+c^3a^3=(ab+bc+ca)^3-3abc(u+v)-6(abc)^2=-4\Longrightarrow uv=-4+3-(a^3+b^3+c^3).$ $a^3+b^3+c^3=(a+b+c)^3-3(u+v)-6abc=8-3+6=11,uv=-12$. Therefore u and v roots equation: $z^2-z-12=(z+1)(z-12)$. Because u>0, we have u=12. $a^2b+b^2c+c^2a=12.$

Rust wrote: $z^2-z-12=(z+1)(z-12)$. Because u>0, we have u=12. $a^2b+b^2c+c^2a=12.$ I think $z^2-z-12=(z-4)(z+3)$ You mistaked too.

Yes $u=4.$

Yes right. It's the same as our "official" solution

since $\sum a^2b+\sum ab^2=(ab+bc+ca)(a+b+c)-3abc=1,\sum a^2b-\sum ab^2=(a-b)(b-c)(a-c)\ge 0$ we only have to compute$|\prod (a-b)|.$ notice that $P'(a)=(a-b)(a-c)$,etc. so$|\prod (a-b)|=\sqrt{|P'(a)P'(b)P'(c)|}=\sqrt{\prod{3a^2-4a-1}}=7$ the last step was achieved by Newton's formula(and brutal force!) hence the vaule is $4$

actually,we can compute $\sum_{sym}x^{2}y$ and $(\sum_{cyc}x^{2})(\sum_{cyc}xy^{2})$ by Newton theorem

Assume the roots of the equation $P(x)=x^3-2x^2-x+1$ are $\alpha,\beta,\gamma.$ By Vieta's Relation we get, $$\alpha+\beta+\gamma=2$$$$\alpha\beta+\beta\gamma+\gamma\alpha=-1$$$$\alpha\beta\gamma=-1$$Given $\alpha>\beta>\gamma\in \mathbb{R}.$ $$\bigg|(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)\bigg|=\bigg|\underbrace{(\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha)}_{=S_1}-\underbrace{(\alpha\beta^2+\beta\gamma^2+\gamma\alpha^2)}_{=S_2}\bigg|$$Now Simply we have to find an equation whose roots are $S_1$ and $S_2.$ Therefore $$S_1+S_2=\sum_{cyc}{}\alpha^2(\beta+\gamma)\implies\sum_{cyc}{}\alpha^2(2-\alpha)\implies 2(\sum_{cyc}{}\alpha^2)-(\sum_{cyc}{}\alpha^3)\implies12-11=\boxed{1}$$$$S_1\cdot S_2=\sum_{cyc}{}\alpha^4\beta\gamma+3(\alpha\beta\gamma)^2+\sum_{cyc}{}\alpha^3\beta^3\implies-6+3+-9=\boxed{-12.}$$Therefore we get, the equation as $$\boxed{x^2-x-12=0=(x-4)(x+3)}$$Therefore the values are $4$ and $-3$. Vieta's Relation!!