Q:

If $x^3 + y^3 + z^3 = 0$, $x$ and $y$ are odd positive integers while $z$ is an even negative integer then $x ≠ y$. How come?

While reading through an article on Wikipedia on the proof of Fermat's Last Theorem for n = 3, I came across an argument that I don't quite understand.
Here's some background on the problem:
Assume that there exists a solution $(x, y, z)$ to the equation $x^3 + y^3 + z^3 = 0$, where the three non-zero integers $x$, $y$, and $z$ are pairwise co-prime and not all positive. One of the three must be even, whereas the other two are odd. Without loss of generality, $z$ may be assumed to be even.
Now comes the line of reasoning that troubles me:

Since $x$ and $y$ are both odd, they cannot be equal. If $x = y$, then $2x^3 = −z^3$, which implies that $x$ is even, a contradiction.

How do you infer from $2x^3=−z^3$ that x is even? As far as I can tell, $-z^3$ has $2$ as one of it's factors. Therefore, $z$ is even as it is divisible by $2$.

A:

If $2x^3=-z^3$, then $z$ must be even, so that $z^3$ is divisible by $8$.
So if $x$ is odd, then $2x^3$ is not divisible by $8$. Hence, $x$ must be even.

A:

You infer it as follows:
$2x^3 = -z^3 \Longrightarrow 2 \mid z^3 \Longrightarrow 2 \mid z \Longrightarrow \exists w \in \Bbb Z, \; z = 2w$
$\Longrightarrow 2x^3 = -8w^3 \Longrightarrow x^3 = -4z^3 \Longrightarrow 2 \mid x^3 \Longrightarrow 2 \mid x. \tag 1$

