Sum of Cubes of Three Indeterminates Minus 3 Times their Product

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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 1

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


Then:

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

$\blacksquare$


Proof 2

Consider the determinant:

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

We have:

\(\displaystyle \Delta\) \(=\) \(\displaystyle x \paren {x^2 - y z} - z \paren {y x - z^2} + y \paren {y^2 - x z}\) Definition of Determinant of Order 3
\(\displaystyle \) \(=\) \(\displaystyle x^3 + y^3 + z^3 - 3 x y z\)


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

\(\displaystyle \Delta\) \(=\) \(\displaystyle \begin {vmatrix} x + y + z & x + y + z & x + y + z \\ y & x & z \\ z & y & x \end {vmatrix}\) Multiple of Row Added to Row of Determinant
\(\displaystyle \) \(=\) \(\displaystyle \paren {x + y + z} \paren {x^2 - y z} - \paren {x + y + z} \paren {y x - z^2} + \paren {x + y + z} \paren {y^2 - x z}\) Definition of Determinant of Order 3
\(\displaystyle \) \(=\) \(\displaystyle \paren {x + y + z} \paren {\paren {x^2 - y z} - \paren {y x - z^2} + \paren {y^2 - x z} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x + y + z}\) \(\divides\) \(\displaystyle \Delta\)


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 $1$:

\(\displaystyle \Delta\) \(=\) \(\displaystyle \begin {vmatrix} x + \omega y + \omega^2 z & \omega x + \omega^2 y + z & \omega^2 x + y + \omega z \\ y & x & z \\ z & y & x \end {vmatrix}\) Multiple of Row Added to Row of Determinant:
\(\displaystyle \) \(=\) \(\displaystyle \begin {vmatrix} x + \omega y + \omega^2 z & \omega \paren {x + \omega y + \omega^2 z} & \omega^2 \paren {x + \omega y + \omega^2 z} \\ y & x & z \\ z & y & x \end {vmatrix}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x + \omega y + \omega^2}\) \(\divides\) \(\displaystyle \Delta\) expanding as above


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

\(\displaystyle \Delta\) \(=\) \(\displaystyle \begin {vmatrix} x + \omega^2 y + \omega z & \omega^2 x + \omega y + z & \omega x + y + \omega^2 z \\ y & x & z \\ z & y & x \end {vmatrix}\) Multiple of Row Added to Row of Determinant
\(\displaystyle \) \(=\) \(\displaystyle \begin {vmatrix} x + \omega^2 y + \omega z & \omega^2 \paren {x + \omega^2 y + \omega z} & \omega^2 \paren {x + \omega^2 y + \omega z} \\ y & x & z \\ z & y & x \end {vmatrix}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x + \omega^2 y + \omega}\) \(\divides\) \(\displaystyle \Delta\)


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.

$\blacksquare$