Sum of Cubes of Three Indeterminates Minus 3 Times their Product/Proof 1
Jump to navigation
Jump to search
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
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity