Let Let a, b and c be the roots of the equation $$x^3 +3x^2-1=0$$Then what is the value of expression $a^2b+b^2c+c^2a$.
I got it done by evaluate the sum and difference of
$a^2b+b^2c+c^2a$ and $ab^2+bc^2+ca^2$ and using Vieta rule, but it was too cumbersome. I wanted another solution which is elegant and concise.
Let $x=\frac{1}{2\cos\alpha}.$
Thus, $$\frac{1}{8\cos^3\alpha}+\frac{3}{4\cos^2\alpha}-1=0$$ or $$8\cos^3\alpha-6\cos\alpha-1=0$$ or $$\cos3\alpha=\frac{1}{2}$$ or $$\alpha=\pm20^{\circ}+120^{\circ}k,$$ where $k\in\mathbb Z,$ which gives the following roots: $$\left\{\frac{1}{2\cos20^{\circ}},-\frac{1}{2\cos40^{\circ}},-\frac{1}{2\cos80^{\circ}}\right\}$$
Can you end it now?
I got that our sum is equal to $3$ or $-6$.
There is also the following way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$, $a^2b+b^2c+c^2a=x$ and $a^2c+b^2a+c^2b=y$.
Thus, $u=-1$, $v^2=0$, $w^3=1$ and $$x+y=9uv^2-3w^3=-3,$$ $$xy=\sum_{cyc}(a^3b^3+a^4bc+a^2b^2c^2)=$$ $$=27v^6-27uv^2w^3+3w^6+27u^3w^3-27uv^2+3w^6+3w^6=27u^3w^3+9w^6=-18,$$ which gives again: $$\{x,y\}=\{3,-6\}$$