This question shows how dividing both sides of an equation by some $f(x)$ may eliminate some solutions, namely $f(x)=0$. Naturally, all examples admit $f(x)=0$ as a solution to prove the point.
I tried to find a simple example of an equation that could be solved by dividing both sides by some $f(x)$ but where $f(x)=0$ was not a solution, and failed miserably. Sure, I can divide both sides of, let's say, $x^2-1=0$ by $x$ ($x=0$ is not a solution), but that doesn't help me solve the equation.
I started wondering if actually the equations that can be solved (or at least simplified) by dividing both sides by some $f(x)$ were precisely those where $f(x)=0$ is a solution: by removing a solution, the division reduces the equation to a simpler form. This is particularly obvious with this example given in that question's accepted answer:
By successively dividing by $x-1$, $x-2$ and so on, the equation becomes simpler as the solutions are removed, until there's no solution left ($1=0$).
However, both the accepted answer and the quote in the question itself say that $f(x)=0$ may be a solution, which I also understand as it may not be one.
So, are there equations where dividing by some $f(x)$ significantly improves the equation resolution, without $f(x)=0$ being a solution?