Find $K=a^2b+b^2c+c^2a$ for roots $a>b>c$ of a cubic.

Question

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.

Answer 1

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 $$K=a^2b+b^2c+c^2a=4.$$

Answer 2

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!