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: The question asks to find an elegant proof of
$$ 1-\frac{x}{zx\!+\!x\!+\!1}-\frac{y}{xy\!+\!y\!+\!1}
 -\frac{z}{yz\!+\!z\!+\!1}= \\
\frac{(xyz-1)^2}
 {(zx\!+\!x\!+\!1)(xy\!+\!y\!+\!1)(yz\!+\!z\!+\!1)}. \tag{1} $$
In order to simplify algebraic manipulation define
$$ X:= zx+x+1,\quad Y:= xy+y+1,\quad Z:= yz+z+1. \tag{2}$$
Move all terms to the same side of the equation and
eliminate all denominators to get
$$ 0 = -X\,Y Z+x\, Y Z+X\,y\, Z+X\,Y z+(1-xyz)^2. \tag{3} $$
Define the polynomial expression
$$ A := X\,Y Z-x\, Y Z-X\,y\, Z-X\,Y z. \tag{4} $$
Proving equation $(3)$ and equation $(1)$ is equivalent to
proving $$ A = (1-xyz)^2.\tag{5} $$

One possible proof is to identity the homogeneous parts of
the degree six polynomial $\,A.\,$
For degrees $0$ and $6$ the only contribution is from the
first term of $\,A\,$ and thus
$$ A_0 = 1, \qquad A_6 = (xyz)^2.  $$
For similar reasons,
$$ A_1 = (x+y+z)-x-y-z=0.  $$
$$ A_5 = (xyz)^2(1/x+1/y+1/z) - (xyz)^2(1/z+1/x+1/y) = 0.$$
$$ A_2 = 2(zx+xy+yz)-x(y+z)-y(x+z)-z(x+y)=0.  $$
$$A_4 = 2xyz (x+y+z)-xyz ((x+y)+(y+z)+(z+x))=0. $$
$$ A_3 \!=\! (4xyz\!+\!z^2x\!+\!x^2y\!+\!y^2z) \\
\!-\!xy(2z\!+\!x)\!-\!yz(2x\!+\!y)\!-\!zx(2y\!+\!z)
\!=\! -2xyz. $$
Putting all the parts together proves equation $(5)$.
This is essentially expanding the polynomial expression
$\,A\,$ and doing the same with $\,(1-xyz)^2.$

Another possible proof is to note that $\,A\,$ is a
polynomial in $\,x,y,z\,$ with maximum degree of each
variable being $2$. There are $3^3=27$ possible
monomials in $\,A.\,$ If equation $(5)$ can be proved
for at least $27$ generic values of $\,x,y,z\,$ then
the equation holds in general. Consider
$$ x=a/b,\;\; y=b/c,\;\; z=c/a,\;\;
x y z = 1, \\ d:=a+b+c,\quad (X,Y,Z) = 
d\Big(\frac1b,\;\frac1c,\;\frac1a\Big), \\
A = \frac{d^2}{abc}(d-a-b-c) = 0 = (1-1)^2. $$
We can choose any nonzero values for $\,a,b,c\,$
which gives more than enough values of $\,x,y,z\,$
which proves equation $(5)$.
A simpler variation of this proof uses
$$ x = y = z =: w,\quad x y z = w^3,\quad X=Y=Z=1+w+w^2 =: W,
\\ A = W^3-3wW^2 = W^2(W-3w) = (W(1-w))^2 = (1-w^3)^2. $$
A: 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 $x=y=z$.
A: Note that (just observe to know it, no need to expand it!)
$\mathrm{LHS}(zx + x + 1)(xy + y + 1)(yz + z + 1) - (xyz - 1)^2$ is a polynomial in $z$ of degree at most $2$.
Thus, if we can find three distinct constants $a, b, c$ such that $\mathrm{LHS} = \mathrm{RHS}$ for $z = a, b, c$ respectively,
then we have $\mathrm{LHS} = \mathrm{RHS}$ for all $x, y, z$.
We choose $z = 1, -1, 0$.
For $z = 1$, $\mathrm{LHS} = 1 - \frac{x}{2x + 1} - \frac{y}{xy + y + 1} - \frac{1}{y + 2}$
and $\mathrm{RHS} = \frac{(xy - 1)^2}{(2x + 1)(xy + y + 1)(y + 2)}$.
Similarly, we only need to check $\mathrm{LHS} = \mathrm{RHS}$ for $y = 0, 1, -1$.
For $z = -1$, $\mathrm{LHS} = 1 - x - \frac{y}{xy + y + 1} - \frac{1}{y}$
and $\mathrm{RHS} = \frac{(xy + 1)^2}{(xy + y + 1)(-y)}$.
Similarly, we only need to check $\mathrm{LHS} = \mathrm{RHS}$ for $y = 1, -1, 2$.
For $z = 0$, $\mathrm{LHS} = 1 - \frac{x}{x+1} - \frac{y}{xy + y + 1}$ and
$\mathrm{RHS} = \frac{1}{(x+1)(xy + y + 1)}$. Verify it directly (or only need to check $\mathrm{LHS} = \mathrm{RHS}$ for $y = 0, -1$).
We are done.
