Calculation in Routh's theorem The proof of Routh's theorem concludes with showing$$1-\frac{x}{zx+x+1}-\frac{y}{xy+y+1}-\frac{z}{yz+z+1}=\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}.$$I seek an elegant "proof from the book" of this, rather than one that involves tedious, potentially error-prone algebra. In particular, it feels like symmetries, degree-counting etc. should make this obvious, rather than an accident where the numerator happens to have a nice factorization. My best approaches are these two:
Option 1
The LHS's first two, three and four terms have respective sums $\frac{zx+1}{zx+x+1}$,$$\frac{(zx+1)(xy+y+1)-y(zx+x+1)}{(zx+x+1)(xy+y+1)}=\frac{x^2yz+zx+1}{(zx+x+1)(xy+y+1)}$$and$$\frac{x^2yz+xz+1}{(zx+x+1)(xy+y+1)}-\frac{z}{yz+z+1}=\frac{(x^2yz+xz+1)(yz+z+1)-z(zx+x+1)(xy+y+1)}{(zx+x+1)(xy+y+1)(yz+z+1)}.$$The numerator is nine monic terms of degree $0$ to $6$ minus nine monic terms of degree $1$ to $5$, so the terms of degree $0$ and $6$ will survive as $(xyz)^2+1$, and any other surviving term(s) will have coefficients summing to $-2$. The problem's symmetries mandate $-2xyz$ to finish the job.
That's quite nice, but the third partial sum probably can't be done in one's head. What one can say, however, is the third partial sum's numerator will be six terms of degrees $0$ to $4$ minus three of $1$ to $3$, so a $0$ and a $4$ survives, but it's harder to deduce without calculation that the third uncancelled term will be of degree $2$.
Option 2
This one looks like it might end up more elegant at first, but it looks like it ultimately requires some of Option 1's techniques to finish.
The case $x=\tfrac{1}{yz}$ has left-hand side$$1-\frac{1}{yz+z+1}-\frac{yz}{yz+z+1}-\frac{z}{yz+z+1}=0,$$so the general case's numerator must be divisible by $xyz-1$. In the special case $x=y=z$, the left-hand side is$$1-\frac{3x}{x^2+x+1}=\frac{(x-1)^2}{x^2+x+1}=\frac{(x^3-1)^2}{(x^2+x+1)^3}.$$The most obvious generalization with appropriate symmetries and denominator is $\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}$, as desired. The most general numerator is of the form$$(xyz-1)(xyz+1+p(x,\,y,\,z)),$$where $p$ is invariant under a cyclic permutation of $x,\,y,\,z$, with $p(x,\,x,\,x)=0$ and $p\left(\tfrac{1}{yz},\,y,\,z\right)=0$.
The first constraint makes $p$ a polynomial in $a:=x+y+z,\,b:=xy+yz+zx,\,c:=xyz$; the second ensures that polynomial vanishes when $a=3x,\,b=3x^2,\,c=x^3$. These are achievable with a factor such as $a^2-3b$, $a^3-27c$, $ab-9c$ or $b^3-27c^2$. The third constraint only adds one requirement, divisibility by $c-1$. Ultimately, some careful degree-counting is needed to prove $p=0$.

A variant on my option 2: as @RiverLi notes, the numerator is of at most degree $2$ in $z$, so is of the form $(c-1)(Aa+Bb+Cc+D)$. That $A=B=0,\,C=D=1$ follows from the case $x=y=z$.
