Name for the identity $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-zy)$?

Question

Within the math literature written in Portuguese and Spanish (at least), the identity

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-zy)$$

is often referred as being "Gauss identity", but I haven't seen this identity with this name in the math literature in English.

Question: is there any base to call this identity "Gauss identity"? is there any specific name for it?

I think it's the Euler's identity. – Michael Rozenberg Oct 15 '18 at 18:19 — Michael Rozenberg, Oct 15 '18 at 18:19
And you can use Newton's identity to prove it... – Dietrich Burde Oct 15 '18 at 18:50 — Dietrich Burde, Oct 15 '18 at 18:50
Identité de Gauss in French. – user376343 Oct 15 '18 at 21:40 — user376343, Oct 15 '18 at 21:40

Will Jagy · Answer 1 · 2018-10-15T21:21:51.937

Not familiar with any name for it.

Note, taking $\omega^3 = 1$ but $\omega \neq 1,$ so that $\omega^2 + \omega = -1,$ $$ x^2 + y^2 + z^2 - yz - zx - xy = (x+y \omega + z \omega^2)(x+y \omega^2 + z \omega) $$

The general rule involved: take your homogeneous ternary cubic. Write down the Hessian matrix of second partial derivatives. The entries of this matrix are linear in the variables. Finally, take the determinant, call that $\mathcal H.$

The theorem is this: the original form, call it $f(x,y,z),$ factors completely into three linear factors (over the complex numbers) if and only if $\mathcal H$ is a constant multiple of $f.$

G. Brookfield:

score 0 · Answer 2 · answered Oct 15 '18 at 19:05

0

There is an article in the American Mathematical Monthly calling this identity Gauss's Gem, see here. So I suppose, Spanish and English have something in common concerning this identity.

answered Oct 15 '18 at 19:05

Dietrich Burde

130,978

Name for the identity $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-zy)$?

2 Answers2