I'm assuming you mean equality not inequality.It doesn't matter where we start since inequality is reflexive.
In order to show ##A \neq B## you need either show ##A \not\subset B## or ##B \not\subset A##. (Or both, but only one is necessary.)Showing a set A does not equal a set B is just as good as showing the set B does not equal the set A.
In this situation ##A \subset B## but ##B \not\subset A##. The original post is making the assumption they should be equal. We are explaining why that is invalid.
You're not disagreeing with what I'm saying.We were done when we found both sets and found the counter example, 0, in the second equation's solution set.
So, we can conclude that (1) x(x^2-1)=0 and (2) x^2-1=0 are different polynomials with different roots. (:
Do note that because (2) is a factor of (1), we can see that (2)'s solutions form a subset less than or equal to (1)'s solutions. More generally, whatever polynomial you multiply (2) by with result in a polynomial whose solution set contains the solutions of (2). This is precisely why we factor polynomials to find their roots. (or wolfram it)
The problem is the original poster did make the assumption that they "should have" the same roots (i.e. that his manipulations were reversible) and ended up in confusion.