Sum of Two Cubes in Complex Domain

Theorem

 * $a^3 + b^3 = \paren {a + b} \paren {a \omega + b \omega^2} \paren {a \omega^2 + b \omega}$

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

Proof
From Sum of Cubes of Three Indeterminates Minus 3 Times their Product:


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

Setting $x \gets 0, y \gets a, z \gets b$:


 * $0^3 + a^3 + b^3 - 3 \times 0 \times a b = \paren {0 + a + b} \paren {0 + \omega a + \omega^2 b} \paren {0 + \omega^2 a + \omega b}$

The result follows.