Let r, s, and t be roots of the equation x^3-5x^2+6x=9. Compute rs/t+st/r+tr/s.

$\dfrac{rs}{t}+\dfrac{st}{r}+\dfrac{tr}{s}\\ =\dfrac{r^2s^2}{rst}+\dfrac{s^2t^2}{rst}+\dfrac{t^2r^2}{rst}\\ =\dfrac{r^2s^2+s^2t^2+t^2r^2}{rst}$

$\text{Let x be }(rs + st + tr)^2\\ x=(rs + st + tr)^2 = r^2s^2+s^2t^2+t^2r^2+2rs^2t+2r^2st+2rst^2\\ \therefore r^2s^2+s^2t^2+t^2r^2=(rs + st + tr)^2 - 2rst(s+r+t)$

$\text{Substitute the expression.} \\\dfrac{r^2s^2+s^2t^2+t^2r^2}{rst}\\=\dfrac{(rs+st+tr)^2-2rst(r+s+t)}{rst}$

$\text{Substitute the values.}\\ r+s+t = \dfrac{-(-5)}{1}=5\\ rst = \dfrac{-(-9)}{1}=9\\ rs+st+tr = \dfrac{6}{1}= 6$

$\dfrac{(rs+st+tr)^2-2rst(r+s+t)}{rst}\\=\dfrac{6^2-2(9)(5)}{9}\\ =4-10\\ =-6$

$\therefore \dfrac{rs}{t}+\dfrac{st}{r}+\dfrac{tr}{s}= -6$