Show that $a\rho b\iff a^4b^2-4b^2=b^4a^2-4a^2$ is equivalence relation and deduce classes

Question

I think I've solved problem, but I need to check it.

Problem: On $\mathbb{R}$ is given relation $a\rho b\iff a^4b^2-4b^2=b^4a^2-4a^2$. Prove that this is equivalence relation and deduce equivalence classes of elements $0$, $\frac{1}{2}$, $-2$.

My solution: I proved that this relation is reflexive, symmetric and transitive and therefore is equivalence relation, but I have to check equivalence classes. I got that $[0]=\{0\}$, $[\frac{1}{2}]=\{\pm \frac{1}{2}\}$, $[-2]=\{\pm 2\}$. Am I right?

Any help is welcome. Thanks in advance.

P.S. Is there any shorter way to conclude that this is equivalence relation and to deduce classes of equivalence?

Answer 1

Hint: Show that $a^4b^2-4b^2=b^4a^2-4a^2 \Leftrightarrow a = \pm b$.

Clearly the latter is an equivalence relation, with elements of the same absolute value.