My friend pass me a simple proof that there is no positive integer solution for equation $a^3+b^3=c^3$.
I'm not sure whether the proof is right or not.
The proof:
suppose $a,b,c$ are positive integers , coprime pairwisely and satisfy the equation $a^3+b^3=c^3$
then we get
$(a+b-c)^3=3(a+b)(c-a)(c-b)~~~~~~(1)$
let $x=a+b$, so $x$ divides $c^3$ ,since $c^3 = a^3+b^3=(a+b)(a^2+ab+c^2)$
so, let $y=c-b,z=c-a$,then $y$ divides $a^3$, then $z$ divides $b^3$
we have $x,y,z$ are pairwisely coprime since $a^3,b^3,c^3$ are pairwisely coprime
by $(1)$, we can see
$(x-y-z)^3=8(a+b-c)^3=24xyz~~~~~(2)$
then $24xyz$ should have factor $x^3$, so $x^2$ divides $24$
Then $x=1$ or $2$
It's imposible.
Your main result says that:
$$(x-y-z)^3=24xyz$$
...whcih basically means that $P=24xyz$ must be a perfect cube. Fair enough, but why does it implicate that $x^3$ divides $P$?
For example, $P$ is a perfect cube for $x=9$, $y=10^3$, $z=11^3$. But $9^3\nmid24\times9\times10^3\times11^3$.