Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find the answer of the following expression, without solving the original equation?
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a}
$$

As I've suggested in the comment yesterday,
let $x=2m\cos y\implies(2m\cos y)^3-(2m\cos y)+1=0\ \ \ \ (1)$
As $\cos3y=4\cos^3y-3\cos y,$
$2m^3(\cos3y+3\cos y)-(2m\cos y)+1=0$
$\iff2m^3\cos3y+2m\cos y(m^2-1)+1=0\  \ \ \ (2)$
WLOG choose $m^2-1=0\iff m=\pm1$
Let $m=1$
$(1)$ reduces to $8\cos^3y-6\cos y+1=0 \ \ \ \ (3)$
and $(2)\implies\cos3y=-\dfrac12\implies3y=360^\circ n\pm120^\circ$ where $n$ is any integer
$\implies y=120^\circ n+40^\circ$ where $n\equiv-1,0,1\pmod3$
So, the roots of $(3)$ are
$\cos(-80^\circ)=\cos80^\circ$ $\cos40^\circ,\cos160^\circ=\cos(180^\circ-20^\circ)=-\cos20^\circ<0$
Clearly, $\cos40^\circ>\cos80^\circ>0>-\cos20^\circ$ 
$\implies c=2\cos40^\circ, b=2\cos80^\circ, a=2\cos160^\circ$
$\implies\dfrac ab=\dfrac{2\cos160^\circ}{2\cos80^\circ}=\dfrac{2\cos^280^\circ-1}{\cos80^\circ}=2\cos80^\circ-\dfrac1{\cos80^\circ}$
$\implies\sum_{\text{cyc}}\dfrac ab=2\sum_{\text{cyc}}\cos80^\circ-\sum_{\text{cyc}}\dfrac1{\cos80^\circ}$
Using Vieta's formula on $(3),\sum_{\text{cyc}}\cos80^\circ=0,$
$\cos40^\circ\cos80^\circ+\cos40^\circ\cos160^\circ+\cos80^\circ\cos160^\circ=\dfrac{-6}8$
and $\cos40^\circ\cos80^\circ\cos160^\circ=-\dfrac18$
and $\sum_{\text{cyc}}\dfrac1{\cos80^\circ}=\dfrac{\cos40^\circ\cos80^\circ+\cos40^\circ\cos160^\circ+\cos80^\circ\cos160^\circ}{\cos40^\circ\cos80^\circ\cos160^\circ}=\cdots=6$
Won't you try with $m=-1?$
