Prove that $$\frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^\pi+e^e}{x-\pi-e}=0$$ has one real root in$(e,\pi)$ and other in $(\pi,\pi+e)$.
For $x\in (-\infty,e)$ the equation will be always be negative and for $x\in (\pi+e,\infty)$ the equation will be always positive so roots must lie in $(e,\pi+e)$.
Now how to do for specific intervals?
Like, if we could prove equation is negative for x=$e^+$ and positive for $x=\pi^-$,then one real root would lie in this interval.
But how do we do this?
The roots of $$f(x)=\frac{\pi^e}{x-e}+\frac{e^\pi}{x-\pi}+\frac{\pi^\pi+e^e}{x-\pi-e}=0$$ and $$g(x)=\pi^e(x-\pi)(x-\pi-e)+e^{\pi}(x-e)(x-e-\pi)+(\pi^\pi+e^e)(x-e)(x-\pi)=0$$ would coincide Note that $$g(e)=\pi^{1+e}(\pi-e)>0,~~ g(\pi)=e^{\pi+1}(\pi-e)<0,~~ g(e+\pi)=e\pi(\pi^\pi+e^e)>0$$ This proves that the quadratic $g(x)$ has one real root in $(e,\pi)$ and other one in $(\pi, \pi+)e.$ So will be the case for $f(x)=0.$