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

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
• And you can use Newton's identity to prove it... – Dietrich Burde Oct 15 '18 at 18:50
• Identité de Gauss in French. – user376343 Oct 15 '18 at 21:40

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 $$f.$$