Find $K=a^2b+b^2c+c^2a$ for roots $a>b>c$ of a cubic. If $a>b>c$ are the roots of the polynomial $P(x)=x^3-2x^2-x+1$ find the value of $K=a^2b+b^2c+c^2a$.
Using Vièta's formulas:
$$a+b+c=2$$
$$ab+bc+ca=-1$$
$$abc=-1$$
Using those I found that 
$$a^2+b^2+c^2=6$$
$$a^3+b^3+c^3=11$$
$$a^2b+b^2c+c^2a+ab^2+bc^2+ca^2=1$$
but I can't separate K from it.
 A: HINT: Consider the discriminant of your polynomial; what can you say about its square root given that $a>b>c$? Hover over the yellow box for a (more) complete solution.

 The discriminant of your polynomial is $\Delta=49$, and because $a>b>c$ you also have
 $$a^2b+b^2c+c^2a-a^2c-b^2a-c^2b=(a-b)(a-c)(b-c)=\sqrt{\Delta}=7.$$
 You have already found that
 $$a^2b+b^2c+c^2a+ab^2+bc^2+ca^2=1,$$
 and from these two it is not hard to see that
 $$K=a^2b+b^2c+c^2a=4.$$

A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $u=\frac{2}{3},$ $v^2=-\frac{1}{3},$ $w^3=-1$ and
$$\sum_{cyc}a^2b=\frac{1}{2}\sum_{cyc}(a^2b+a^2c+a^2b-a^2c)=$$
$$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}(a-b)(a-c)(b-c)=$$
$$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}\sqrt{(a-b)^2(a-c)^2(b-c)^2}=$$
$$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}\sqrt{27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)}=$$
$$=\frac{1}{2}\left(9\cdot\frac{2}{3}\left(-\frac{1}{3}\right)-3(-1)\right)+$$
$$+\frac{1}{2}\sqrt{27\left(3\left(\frac{2}{3}\right)^2\left(-\frac{1}{3}\right)^2-4\left(-\frac{1}{3}\right)^3-4\left(\frac{2}{3}\right)^3(-1)+6\cdot\frac{2}{3}\left(-\frac{1}{3}\right)(-1)-(-1)^2\right)}=4.$$
The hint for another way:
Prove that
$$(a,b,c)=\left(1+2\cos\frac{2\pi}{7},1+2\cos\frac{4\pi}{7},1+2\cos\frac{6\pi}{7}\right)$$ and end it!
