How can I find the value of a cyclic sum whose arguments are known to be roots to a polynomial?
The sum I want to find is
$$\sum_{\mathrm{cyc}}\frac{1}{xy+z-1}=\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{zx+y-1}$$
I was given a set of power/symmetric sums that allowed me to use Vieta's formulas to determine that the numbers $x,y,z$ are roots to the cubic $t^3-2t^2+\frac{1}{2}t-4$ (verified with Mathematica). Are there any relations between this cubic's roots that I can rely on to find the value of the sum? (The answer should be $-\frac{2}{9}$.)
I've thought to combine the fractions, but expanding the resulting products and appropriately collecting terms is rather tedious. True, some terms reduce to known quantities, such as the leading term in the numerator ($xyz^2+xy^2z+x^2yz=xyz(x+y+z)=8$), but other terms don't seem so convenient.
Note that $x+y+z$ is the opposite of the coefficient of $t^2$ in the polynomial $$P(t)=t^3−2t^2+\frac12t−4$$ hence $$x+y+z=2$$ which implies $$xy+z-1=xy-x-y+1=(x-1)(y-1)$$ Thus, the cyclic sum $S_{\mathrm{cyc}}$ to be computed is also $$S_{\mathrm{cyc}}=\sum_{\mathrm{cyc}}\frac{z-1}{(x-1)(y-1)(z-1)}=\frac{x+y+z-3}{(x-1)(y-1)(z-1)}=\frac{-1}{(x-1)(y-1)(z-1)}$$ where the last identity uses again the value of $x+y+z$. Now, $x-1$, $y-1$ and $z-1$ are the roots of the polynomial $$ P(t+1)=(t+1)^3−2(t+1)^2+\frac12(t+1)−4$$ and, as such, their product $(x-1)(y-1)(z-1)$ is the opposite of the constant term of $P(t+1)$, which is the value of $P(t)$ at $t=1$, that is, $$-(x-1)(y-1)(z-1)=P(1)=1−2+\frac12−4=-\frac92$$ which proves indeed that $$S_{\mathrm{cyc}}=\frac1{P(1)}=-\frac29$$