<s>[INST] 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.
 [/INST] 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
 $$K=a^2b+b^2c+c^2a=4.$$

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