Question about difference of powers related to FLT I am wondering about differences of powers and some idea about it. It is related to FLT in that if the idea is correct something good could be done with it. I will give an example for prime $3$ but it could easily work for any odd prime. My question is: is the idea correct?
Let the following equation hold:
$$x^3 - y^3 = 3y^2(x - y) + 3y(x - y)^2 + (x - y)^3 = z^3$$
for some integers $x,y,z$.
We can obtain this by means of the binomial expansion of $((x - y) + y)^3 = x^3$ and subtracting $y^3$ to get $z^3$.
Rewrite the equation as follows(by recursively factoring out $x - y$):
$$(x - y)(3y^2 + (x - y)(3y + (x - y))) = z^3$$
We will prove: $p \mid x - y \implies p^3 \mid x - y$ for any prime other than $3$:
Assume $\gcd(x,y,z) = 1$,
Let $p \ne 3$ be some prime  for which $p \mid x - y$, 
$\quad\quad p \mid x - y \implies p \mid z \implies p^3 \mid z^3$
So we get:
$\quad\quad (x - y)(3y^2 + (x - y)(3y + (x - y))) \equiv z^3 \equiv 0 \pmod{p^3}$
Suppose now $p^3 \nmid x - y$, then:
$\quad\quad 3y^2 + (x - y)(3y + (x - y)) \equiv 0 \pmod{p}$
And so:
$\quad\quad (x - y)(3y + (x - y)) \equiv -3y^2 \pmod{p}$
$\quad\quad \implies p \mid 3y^2$ which is impossible because of $p \ne 3$ and $\gcd(x,y,z) = 1$.
We may conclude that for every prime $p \ne 3$, $p \mid x - y$ we have $p^3 \mid x - y$
The same idea should work for prime-divisors of $x - z$.

Assuming we have $z^3=x^3-y^3=(x-y)(x^2+xy+y^2)$ and then
\begin{align}
\gcd(x-y,x^2+xy+y^2) &= \gcd(x-y,x^2+xy+y^2)\\
&= \gcd(x-y,2xy+y^2)\\
&= \gcd(x-y,3xy).
\end{align}
But because you assume also $\gcd(x,y,z)=1$, it follows $\gcd(x-y,x^2+xy+y^2)=\gcd(x-y,3)$.
In other words, $x-y$ and $x^2+xy+y^2$ have no common factor apart from $3$, and so of course, if you find prime $p \mid x-y$, $p\neq 3$ then indeed $p^3 | x-y$. 
But that is not new, same applies for generic $n$. If you have 
$$z^n=x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1})$$ then you can show similarly as before (assuming $\gcd(x,y,z)=1$ again) that
$$
\gcd(x-y,x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1})=\gcd(x-y,n)
$$
and so if there is a prime $p$ such that $p\nmid n$ and $p|x-y$, then it follows $p^n\mid x-y$.
So the idea is correct, question is whether it is of any use, which I do not know.
