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

*

*Reflexive: 
$m^3 +m = m^3 +m$


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


*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?

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