This is one of those recreational questions that aren't really about research. I found a curious remark in an old volume of American Mathematical Monthly (1922) which I'll quote below:

In the Proceedings of the Royal Society of Edinburgh, vol.7, p.144, in some mathematical notes by professor P.G. Tait, it is stated:

 

"If $x^3+y^3=z^3$, then $(x^3+z^3)^3y^3+(x^3-y^3)^3z^3=(z^3+y^3)^3x^3$.

 

This furnishes an easy proof of the impossibility of finding two integers the sum of whose cubes is a cube."

 

How does this "easy proof" follow? Students are notoriously suspicious of those steps which an author announces as "easy", and are sometimes inclined to believe that the word is used in a humorous sense. [...] There are of course proofs in existence that the sum of two cubes cannot be a cube.

Did anyone manage to find a proof of FLT for the exponent 3 using this identity or is the alluded proof another illusion that did not fit in the margin?

This is one of those recreational questions that aren't really about research. I found a curious remark in an old volume of American Mathematical Monthly (1922) which I'll quote below:

In the Proceedings of the Royal Society of Edinburgh, vol.7, p.144, in some mathematical notes by professor P.G. Tait, it is stated:

"If $x^3+y^3=z^3$, then $(x^3+z^3)^3y^3+(x^3-y^3)^3z^3=(z^3+y^3)^3x^3$.

This furnishes an easy proof of the impossibility of finding two integers the sum of whose cubes is a cube."

How does this "easy proof" follow? Students are notoriously suspicious of those steps which an author announces as "easy", and are sometimes inclined to believe that the word is used in a humorous sense. [...] There are of course proofs in existence that the sum of two cubes cannot be a cube.

Did anyone manage to find a proof of FLT for the exponent 3 using this identity or is the alluded proof another illusion that did not fit in the margin?