Using Vieta's formula to evaluate $\frac{x_1}{x_2} + \frac{x_2}{x_3} + \frac{x_3}{x_1}$ Using the cubic equation $ax^3 + bx^2 + cx +d = 0$,
$$x_1 + x_2 + x_3 = -\frac{b}{a},$$
$$x_1x_2 + x_2x_3 + x_1x_3 = \frac{c}{a},$$
$$x_1x_2x_3 = -\frac{d}{a},$$
How would one evaluate $\frac{x_1}{x_2} + \frac{x_2}{x_3} + \frac{x_3}{x_1}$?
Edit: I'm at this point currently:
$$\frac{x_1^2x_3 + x_1x_2^2 + x_2x_3^2}{x_1x_2x_3}$$
and don't know how to separate the single $x_1, x_2, x_3$ from the fraction.
 A: Let $r,s,t$ be the roots.  For simplicity, write $B=-b/a=r+s+t$, $C=c/a=st+tr+rs$, and $D=-d/a=rst$.  Let $p= \frac{r}{s}+\frac{s}{t}+\frac{t}{r}$ and $q=\frac{s}{r}+\frac{t}{s}+\frac{r}{t}$.  Then,
$$p+q=\frac{r^2(s+t)+s^2(t+r)+t^2(r+s)}{rst}.$$
$$pq=3+\frac{r^4st+s^4tr+t^4rs+s^3t^3+t^3r^3+r^3s^3}{r^2s^2t^2}.$$
That is,
$$p+q=\frac{BC-3D}{D}$$
and
$$pq=3+\frac{C^3+B^3D-6BCD+6D^2}{D^2}=\frac{C^3+B^3D-6BCD+9D^2}{D^2}.$$
Therefore, $p$ and $q$ are the roots of
$$x^2-\frac{BC-3D}{D}x+\frac{C^3+B^3D-6BCD+9D^2}{D^2},$$
which is the same as
$$x^2+\frac{3ad-bc}{ad}x+\frac{9a^2d^2-6abcd+ac^3+b^3d}{a^2d^2}.$$
For example, $B=6$, $C=11$, and $D=6$ give $$x^2-8x+\frac{575}{36}=\left(x-\frac{23}{6}\right)\left(x-\frac{25}{6}\right).$$
Indeed, $\{r,s,t\}=\{1,2,3\}$, so $$\{p,q\}=\left\{\frac12+\frac23+\frac31,\frac21+\frac32+\frac13\right\}=\left\{\frac{25}{6},\frac{23}{6}\right\}.$$
A: You can consider $$\frac{x_2}{x_1} + \frac{x_3}{x_2} + \frac{x_1}{x_3}$$
The sum and product of this and your expression are symmetric. So you can evaluate them using $a, b, c, d$. 
So this things are roots of the polynomial of degree 2 with coefficients expressed by $a, b, c, d$.  
