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
\(\ds \) | \(\) | \(\ds \paren {x + \omega y + \omega^2 z} \paren {x + \omega^2 y + \omega z}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^2 + \omega^2 x y + \omega x z + \omega x y + \omega^3 y^2 + \omega^2 y z + x \omega^2 z + \omega^4 y z + \omega^3 z^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^2 + y^2 + z^2 + \paren {\omega + \omega^2} x y + \paren {\omega + \omega^2} x z + \paren {\omega + \omega^2} y z\) | $\omega^3 = 1$ 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:
\(\ds \) | \(\) | \(\ds \paren {x + y + z} \paren {x^2 + y^2 + z^2 + - x y - x z - y z}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^3 + x y^2 + x z^2 - x^2 y - x^2 z - x y z\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds y x^2 + y^3 + y z^2 - x y^2 - x y z - y^2 z\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds z x^2 + z y^2 + z^3 - x y x - x z^2 - y z^2\) | |||||||||||
\(\ds \) | \(=\) | \(\ds x^3 + y^3 + z^3 - 3 x y z\) |
$\blacksquare$
Proof 2
Consider the determinant:
- $\Delta = \begin {vmatrix} x & z & y \\ y & x & z \\ z & y & x \end {vmatrix}$
We have:
\(\ds \Delta\) | \(=\) | \(\ds x \paren {x^2 - y z} - z \paren {y x - z^2} + y \paren {y^2 - x z}\) | Determinant of Order 3 | |||||||||||
\(\ds \) | \(=\) | \(\ds x^3 + y^3 + z^3 - 3 x y z\) |
Then we note that adding rows $2$ and $3$ to rows $1$ gives:
\(\ds \Delta\) | \(=\) | \(\ds \begin {vmatrix} x + y + z & x + y + z & x + y + z \\ y & x & z \\ z & y & x \end {vmatrix}\) | Multiple of Row Added to Row of Determinant | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x + y + z} \paren {x^2 - y z} - \paren {x + y + z} \paren {y x - z^2} + \paren {x + y + z} \paren {y^2 - x z}\) | Determinant of Order 3 | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x + y + z} \paren {\paren {x^2 - y z} - \paren {y x - z^2} + \paren {y^2 - x z} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x + y + z}\) | \(\divides\) | \(\ds \Delta\) |
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 $1$:
\(\ds \Delta\) | \(=\) | \(\ds \begin {vmatrix} x + \omega y + \omega^2 z & \omega x + \omega^2 y + z & \omega^2 x + y + \omega z \\ y & x & z \\ z & y & x \end {vmatrix}\) | Multiple of Row Added to Row of Determinant: | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin {vmatrix} x + \omega y + \omega^2 z & \omega \paren {x + \omega y + \omega^2 z} & \omega^2 \paren {x + \omega y + \omega^2 z} \\ y & x & z \\ z & y & x \end {vmatrix}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x + \omega y + \omega^2}\) | \(\divides\) | \(\ds \Delta\) | expanding as above |
and adding $\omega^2$ times row $2$ and $\omega$ times row $3$ to rows $1$:
\(\ds \Delta\) | \(=\) | \(\ds \begin {vmatrix} x + \omega^2 y + \omega z & \omega^2 x + \omega y + z & \omega x + y + \omega^2 z \\ y & x & z \\ z & y & x \end {vmatrix}\) | Multiple of Row Added to Row of Determinant | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin {vmatrix} x + \omega^2 y + \omega z & \omega^2 \paren {x + \omega^2 y + \omega z} & \omega^2 \paren {x + \omega^2 y + \omega z} \\ y & x & z \\ z & y & x \end {vmatrix}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x + \omega^2 y + \omega}\) | \(\divides\) | \(\ds \Delta\) |
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.
$\blacksquare$