If α, β and γ are the zeroes of $x^3-9x^2+23x-15$, the value of 1/(αβ) + 1/(βγ) + 1/(γα) is:

Put all three over the common denominator $\alpha \beta \gamma :\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha }=\frac{\gamma +\alpha +\beta }{\alpha \beta \gamma }$. Here $\alpha +\beta +\gamma =-\frac{b}{a}=9$ and $\alpha \beta \gamma =-\frac{d}{a}=15$, so the value is $\frac{9}{15}=\frac{3}{5}$. (The zeroes are 1, 3 and 5.) $\frac{5}{3}$ inverts it; $\frac{23}{15}$ puts the pairwise sum 23 on top, which answers $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }$ rather than this; $\frac{9}{23}$ divides by the wrong coefficient.