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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?

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

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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: enter image description here

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