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?
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a}
$$

We know that every symmetric function of the roots $a,b,c$ can be evaluated in terms of the elementary symmetric functions:
$$ e_1=a+b+c=0,\quad e_2=ab+ac+bc=-3,\quad e_3=abc=-1$$
or the power sums:
$$ p_1=e_1=0,\quad p_2=a^2+b^2+c^2 = 6,\quad p_3=a^3+b^3+c^3=3e_1-3=-3.$$
Now:
$$ g(a,b,c)=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=-(a^2 c+b^2 a+c^2 b)$$
is not a symmetric function of $a,b,c$, and neither it is:
$$ h(a,b,c)=\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=-(a^2 b+b^2 c+c^2 a),$$
but both $g+h$ and $g\cdot h$ are. So the strategy is just to find $g+h$ and $g\cdot h$ in terms of $e_1,e_2,e_3$, then solve a quadratic equation to find $\{g,h\}$ and recognize $g$ from the constraint $a<b<c$.
We have:
$$\begin{eqnarray*} g+h &=& -(a^2(b+c)+b^2(a+c)+c^2(a+b))\\ &=& (a^3+b^3+c^3)-(a^2+b^2+c^2)(a+b+c)\\&=&p_3-p_2 p_1=-3,\end{eqnarray*}$$
$$\begin{eqnarray*} g\cdot h &=& e_3 p_3 + 3a^2b^2c^2 + e_3^3\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\\&=&6-\left(\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}-3\right)\\&=&9-3\left(\frac{e_2^2}{e_3^2}-2\frac{e_1}{e_3}\right)=-18,\end{eqnarray*}$$
hence $g,h$ are the roots of $z^2+3z-18$, and $\{g,h\}=\{-6,3\}$. Since $e_3<0$, we have $a<0<b<c$, from which:
$$ -g = a^2 c+b^2 a+ c^2 b = (b+c)^2 c-b^2(b+c)+c^2 b = c^3-b^3+3bc^2 > 0 $$
and $\color{red}{g=-6}$ follows.
