Sum of Cubes of Three Indeterminates Minus 3 Times their Product/Proof 2

Proof
Consider the determinant:


 * $\Delta = \begin {vmatrix} x & z & y \\ y & x & z \\ z & y & x \end {vmatrix}$

We have:

Then we note that adding rows $2$ and $3$ to rows $1$ gives:

Let $\omega$ denote the complex cube root of unity:
 * $\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$

Hence adding $\omega$ times row $2$ and $\omega^2$ times row $3$ to rows $1$:

and adding $\omega^2$ times row $2$ and $\omega$ times row $3$ to rows $1$:

Thus we have $3$ divisors of $x^3 + y^3 + z^3 - 3 x y z$, which is a polynomial of degree $3$.

There can be no other divisors except for a constant.

By examining the coefficient of $x^3$, for example, the constant is seen to be $1$.

Hence the result.