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 have 
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a} = \frac{a^2c+b^2a+c^2b}{abc}
$$
Since from Vieta's relations we know
$$
a+b+c =0,\quad ab+bc+ca =-3,\quad abc =-1,
$$
our goal is to calculate
$$
s = a^2c+b^2a+c^2b.
$$
Let's introduce
$$
p = ac^2+ba^2+cb^2.
$$
Than we have
$$
0 = (ab+bc+ac)(a+b+c) = p+s+3abc 
$$
and $p+s = 3$.
Now let's multiply
$$
s\cdot p = a^3b^3 + a^3c^3 +b^3c^3 + 3(abc)^2+ abc(a^3+b^3+c^3)
$$
Since $a,b,c$ are the roots of polynomial the last equation can be rewritten as
$$
sp = (3a-1)(3b-1)+(3a-1)(3c-1)+(3b-1)(3c-1) + 3(abc)^2
+ abc(3(a+b+c)-3)= 
$$
$$
=9(ab+ac+bc)-6(a+b+c)+3 + 3(abc)^2 + abc(3(a+b+c)-3) =-27+3+3+3=-18.
$$
So, $s+p=3$ and $sp=-18$. From here one can deduce that $s= 6$. (see @mathlove answer)
