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I tried to prove Routh's theorem from geometry and while solving it I had to simpily $$1-\frac{s}{st+s+1}-\frac{t}{rt+t+1}-\frac{r}{rs+r+1}$$ to $$\frac{(rst-1)^2}{(st+s+1)(rt+t+1)(rs+r+1)}$$ I managed to do it by multiplying everything out and cancelling many terms but is there any "clever" way to see the identity?

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Note the symmetry: in each of the subtracted terms, the pattern from left to right (with first component in each parentheses being the "major" and second being the "minor" variable) is $(s,t) \rightarrow (t,r) \rightarrow (r,s)$, which is cyclic. So you would look only at terms of the form: $r+s+t,rst,rs+st+tr$ etc. The symmetry hints to look at what happens if you set $r=s=t=x$, say. You can deduce a lot about the form involving $r,s,t$ from what you get from the $x$ analogue - probably get $\dfrac{(x^3-1)^2}{(x^2+x+1)^3}$. Just need care in translating this back to $r,s,t$ symbols. – Marconius Jul 20 '15 at 23:02

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