Let $$f(x)=8-\binom{4}{0}\frac{1}{x+5}-\binom{4}{1}\frac{1}{x+3}-\binom{4}{2}\frac{1}{x+1}-\binom{4}{3}\frac{1}{x-1}-\binom{4}{4}\frac{1}{x-3}$$ and $$g(x)=8+\binom{4}{0}\frac{1}{x-5}+\binom{4}{1}\frac{1}{x-3}+\binom{4}{2}\frac{1}{x-1}+\binom{4}{3}\frac{1}{x+1}+\binom{4}{4}\frac{1}{x+3}.$$ Show that there are no common roots for $f(x)=0$ and $g(x)=0$.
My approach is to use the intermediate value theorem. By looking at the graph of $f(x)$, we can find $5$ distinct intervals each of which contains exactly one root for $f(x)$. Similarly, we can find them for $g(x)$. Finally, I was able to finish the proof by showing that any two intervals among the $10$ distinct intervals do not overlap.
I wonder if there are any other ways to prove this.
Here's a mechanical way that doesn't require any cleverness, although it does require one tool that you might not have seen yet.
It suffices to show that the polynomials \begin{align*} p(x) &= (x-5) (x-3) (x-1) (x+1) (x+3) (x+5) f(x) \\ &= 8 x^6-16 x^5-264 x^4+480 x^3+1720 x^2-2384 x-1080 \\ q(x) &= (x-5) (x-3) (x-1) (x+1) (x+3) (x+5) g(x) \\ &= 8 x^6+16 x^5-264 x^4-480 x^3+1720 x^2+2384 x-1080 \end{align*} have no common roots. One can then compute the resultant of the two polynomials, which turns out to be $$ \mathop{\rm res}(p,q) = -65{,}355{,}707{,}448{,}508{,}518{,}802{,}391{,}040; $$ since the resultant is nonzero, the two polynomials have no common roots.