Sum of Cubes of Three Indeterminates Minus 3 Times their Product/Proof 1

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Theorem

For indeterminates $x, y, z$:

$x^3 + y^3 + z^3 - 3 x y z = \paren {x + y + z} \paren {x + \omega y + \omega^2 z} \paren {x + \omega^2 y + \omega z}$

where $\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$


Proof

\(\ds \) \(\) \(\ds \paren {x + \omega y + \omega^2 z} \paren {x + \omega^2 y + \omega z}\)
\(\ds \) \(=\) \(\ds x^2 + \omega^2 x y + \omega x z + \omega x y + \omega^3 y^2 + \omega^2 y z + x \omega^2 z + \omega^4 y z + \omega^3 z^2\)
\(\ds \) \(=\) \(\ds x^2 + y^2 + z^2 + \paren {\omega + \omega^2} x y + \paren {\omega + \omega^2} x z + \paren {\omega + \omega^2} y z\) $\omega^3 = 1$ by definition
\(\ds \) \(=\) \(\ds x^2 + y^2 + z^2 - x y - x z - y z\) Sum of Cube Roots of Unity: $\omega + \omega^2 = -1$


Then:

\(\ds \) \(\) \(\ds \paren {x + y + z} \paren {x^2 + y^2 + z^2 + - x y - x z - y z}\)
\(\ds \) \(=\) \(\ds x^3 + x y^2 + x z^2 - x^2 y - x^2 z - x y z\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds y x^2 + y^3 + y z^2 - x y^2 - x y z - y^2 z\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds z x^2 + z y^2 + z^3 - x y x - x z^2 - y z^2\)
\(\ds \) \(=\) \(\ds x^3 + y^3 + z^3 - 3 x y z\)

$\blacksquare$


Sources