<s>[INST] Is the conclusion wrong in this proof of Case 2 of FLT for n=3? The classic proofs of some cases of Fermat´s last theorem are separated into two parts. These proofs use the reduced Fermat equation $z^n=x^n+y^n$.  The positive integers $x,y,z $ are coprime to each other. In Case$1$ they are coprime to the positive integer n, but in Case $2$, one of $x, y, z$ is divisible by $n$.
This attempt to prove Case $2$ for $n=3$ is inspired by the Theorems on FLT by the Norwegian mathematician Niels Henrik Abel. The proof is one of contradiction. The crucial point is the conclusion in the end of the proof.
Assume that (I) $z^3=x^3+y^3$ is true, when the positive integers $x,y,z $ are coprime to each other. Assume $z$ is the one which is divisible by $3$. Factorize (I) $z^3=x^3+y^3$ $\implies $ (II) $z^3=(x+y)((x+y)^2-3xy)$. The left side has $3$ factors $=3$. $ (x+y)^2-3xy$ has only $1$ factor $=3$ since $3xy$ has only $1$ factor $=3$ $\implies$ $x+y $  has $2$ factors $=3$.
Also, $x+y$ divides $z^3$ $\implies$ $x+y$  is coprime to $x$ and to $y$ $\implies$
If $x+y$  is not divisible by $3$, $x+y$  is coprime to $ (x+y)^2-3xy$.
Then $3$ is the only possible common factor between $x+y$ and $ (x+y)^2-3xy$.
$z^3=x^3+y^3$ $\iff $ $y^3=z^3-x^3$ $\iff $ $x^3=z^3-y^3$.
Factorizing of  $y^3=z^3-x^3$ and $x^3=z^3-y^3$ $\implies$
$y^3=(z-x)((z-x)^2+3xz)$  and  $x^3=(z-y)((z-y)^2+3yz)$.
Since $x$ and $y$ are not divisible by $3$ , the factors of these factorizations are coprimes and therefore cubes.
Let $z-x=b^3$ and $z-y=d^3$.  $b,d$ are positive coprime integers. Then
$ b^3+d^3=(z-x) + (z-y) =2z- (x+y)$ $\implies$ (III) $ b^3+d^3=2z- (x+y) $.
The right side has $1$ factor $=3$ $\implies$ the left side is divisible by $3$.
Factorizing of the left side $\implies$ $ b^3+d^3=(b+d)((b+d)^2-3bd)$. Then,
If one factor in the left side is divisible by $3$, the other factor is too $\implies$
The left side has $2$ factors $=3$ since $x+y$ has 2 factors $=3$ but $z$ only $1$ factor. That is a contradiction. $z^3\neq x^3+y^3$.
If $z$ has $2$ factors $=3$ $\implies$ $x+y $  in (II) $z^3=(x+y)((x+y)^2-3xy)$
has $5$ factors $=3$. That does not give a contradiction in (III) $ b^3+d^3=2z- (x+y)$.
But a contradiction is already reached for $z$ divisible by $1$ factor $=3$.
It is shown that $z^3\neq x^3+y^3$ $\iff $ $z^3\neq(x+y)((x+y)^2-3xy)$ when $z$ is divisible by $1$ factor $=3$. Then $z^3\neq(x+y)((x+y)^2-3xy)$ when $z$ is divisible by $2$ factors$=3$. If left side does not equal right side when both sides of the equation  is divided by $3$ factors $=3$, division of both sides by more factors $=3$ does not make them equal.
In the same way, inequality is shown when $x$ or $y$ divisible by $3$.
Case $2$ of FLT for n=3 is true.
 [/INST] The proof is not correct. In line -10 you have used the factorisation of $b^3+d^3$ to prove that this expression is divisible by $9$. You then correctly deduce that $z$ is divisible by $9$.
You then claim that this is a contradiction but it isn't a contradiction, it just means that $9|z$ and $3^5|x+y$.
FLT for $n=3$ can be proved by elementary means - see for example Fermat's Affirmative Questions, The Mathematical Gazette, Vol 96, Number 535, March 2012.
The sort of equations you were using were used very successfully and with great ingenuity by Sophie Germain, for quite general values of $n$, but only for Case 1 of FLT.
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