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

## 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

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.

$\blacksquare$