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.

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity: Exercise $5 \ \text {(i)}$