Proving equivalence relation $a∼b \iff \left(a^2-b^2\right)\left(a^2b^2-1\right)=0$

Question

I don't know how to go about proving symmetry. I have proven that the relation is reflexive. But I have no idea how to start with proving the symmetry of a given relation.

Answer 1

$$(a^2-b^2)(a^2b^2-1)=0\iff (b^2-a^2)(b^2a^2-1)=0.$$

Answer 2

Considering $$y=\left(a^2-b^2\right)\left(a^2b^2-1\right)$$ let $b=k a$ to get $$y=a^2 \left(1-k^2\right) \left(a^4 k^2-1\right)$$ Expand it as a polynomial in $(k-1)$ to get $$y=\left(2 a^2-2 a^6\right) (k-1)+\left(a^2-5 a^6\right) (k-1)^2-4 a^6 (k-1)^3-a^6 (k-1)^4$$ Now, if $b \sim a$, $k\sim 1$.