Showing that $x^3 + y = y^3 + x$ is an equivalence relation

Question

I am asked to prove that: $x^3 + y = y^3 + x$ is an equivalence relation. So far I have the following:

1. Reflexive:
   $m^3 +m = m^3 +m$

2. Symmetric:
   $m^3 + n = n^3 + m \rightarrow n^3 + m = m^3 + n$ Then:
   $n^3 + m = m^3 +n$ From hypothesis

3. Transitivity (here's where I got stuck):
   $m^3 + n = n^3 + m \wedge n^3 + o = o^3 + n \rightarrow m^3 + o = o^3 + m$ Then:
   $m^3 + o = n^3 + m - n = n^3 - n + m = o^3 + n - o - n + m = o^3 - o +m$

And I cant figure out a way to go from $o^3 - o + m$ to $o^3 + m$; what could I do? Am I missing something?

Answer 1

Alternative approach:

$x \sim y \iff x^3 + y = y^3 + x \iff (x^3 - x) = (y^3 - y).$

Then, $\{ ~x \sim y ~~~~\text{and}~~~~ y \sim z ~\} \implies $

$(x^3 - x) = (y^3 - y) ~~~~\text{and}~~~~ (y^3 - y) = (z^3 - z).$

This implies that $(x^3 - x) = (z^3 - z) \iff x \sim z.$

Answer 2

I missed an o on the first step: It should look like this:

1. Transitivy: $m^3 + o = n^3 + m -n + \textbf{o} = n^3 -n + m + \textbf{o} = o^3 +n - o -n +m + \textbf{o} = o^3 + m$