Is the following fact significant? (the factorisation of A^3 +/- B^3)

[Terry Coates]

In the factorisation of A^3 +/- B^3 = (A+/B)(A^2-/+A.B + B^2) the larger factor is the solution for finding the third side of a triangle with sides A and B with the angle between them of 60/120 degrees.(Cosine rule) For any set of three different numbers X>Y>Z there would be three different triangles with the length of the third side being (Y^3 + Z^3)^0.5 (60 deg), (X^3 - Y^3)^0.5 (120 deg) and (X^3 - Z^3)^0.5 (120 deg) Might be of interest in the proof of FLT for power 3.

 

[Charles Link]

By FLT, I presume you mean Fermat's Last Theorem. I will need to study your post further. Perhaps others might have an input.

 

[Mark44]

In the factorisation of A^3 +/- B^3 = (A+/B)(A^2-/+A.B + B^2)

Presumably you intended the first factor on the right to be (A +/- B). You can also write this more clearly as (A ± B), using the symbols under the $\Sigma$ icon on the menu bar.

the larger factor is the solution for finding the third side of a triangle with sides A and B with the angle between them of 60/120 degrees.(Cosine rule)

"angle between them of 60 degrees or 120 degrees"

For any set of three different numbers X>Y>Z there would be three different triangles with the length of the third side being (Y^3 + Z^3)^0.5 (60 deg), (X^3 - Y^3)^0.5 (120 deg) and (X^3 - Z^3)^0.5 (120 deg)

How so? If the two sides used in the Cosine Rule are A and B, and the included angle is either 60° or 120° (with cosines of 1/2 and -1/2, respectively), the length of the third side is $C = \sqrt{A^2 \pm AB + B^2}$.

What you apparently are saying is that $C = \sqrt{A^3 \pm B^3}$, which isn't true, since the right side is $\sqrt{(A \pm B)(A^2 \pm AB + B^2)}$ and you have omitted the $(A \pm B)$ factor in your work with the Cosine Rule above.