<s>[INST] 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?
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a}
$$
 [/INST] By Vieta's formulas, we have
$$a+b+c=-\frac{0}{1}=0\tag1$$
$$ab+bc+ca=\frac{-3}{1}=-3\tag2$$
$$abc=-\frac{1}{1}=-1\tag3$$
From $(1)(2)(3)$, we have
$$P=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2=(a+b+c)(ab+bc+ca)-3abc=3\tag4$$
Now, set
$$Q=\frac ab+\frac bc+\frac ca,\ \ \ R=\frac ba+\frac cb+\frac ac.$$
Then, we have
$$Q+R=\frac ab+\frac bc+\frac ca+\frac ba+\frac cb+\frac ac=\frac{P}{abc}=-3\tag5$$
and
$$\begin{align}QR&=\left(\frac ab+\frac bc+\frac ca\right)\left(\frac ba+\frac cb+\frac ac\right)\\&=3+\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}+\frac{c^2}{ab}+\frac{a^2}{bc}+\frac{b^2}{ca}\\&=3-\frac{1}{a^3}-\frac{1}{b^3}-\frac{1}{c^3}+\frac{a^3+b^3+c^3}{abc}\\&=3-\frac{1}{3a-1}-\frac{1}{3b-1}-\frac{1}{3c-1}+\frac{3abc+(a+b+c)((a+b+c)^2-3(ab+bc+ca))}{abc}\\&\small=3+\frac{-9(ab+bc+ca)+6(a+b+c)-3}{27abc-9(ab+bc+ca)+3(a+b+c)-1}+\frac{3abc+(a+b+c)((a+b+c)^2-3(ab+bc+ca))}{abc}\\&=-18\tag6\end{align}$$
Also, since it is easy to see
$$a\lt 0\lt b\lt c,$$
we have 
$$\frac ab\lt 0,\frac bc\lt 1,\frac ca\lt 0\Rightarrow Q\lt 1\tag 7$$
So, as a result, from $(5)(6)(7)$ we have
$$\color{red}{\frac ab+\frac bc+\frac ca=Q=-6}$$
</s>