Three Times Sum of Cubes of Three Indeterminates Plus 6 Times their Product

Theorem

 * $3 \paren {a^3 + b^3 + c^3 + 6 a b c} = \paren {a + b + c}^3 + \paren {a + b \omega + c \omega^2}^3 + \paren {a + b \omega^2 + c \omega}^3$

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

Proof
Multiplying out:

Replacing $b$ with $b \omega$ and $c$ with $c \omega^2$ in $(1)$:

Replacing $b$ with $b \omega^2$ and $c$ with $c \omega$ in $(1)$:

Adding together $(1)$, $(2)$ and $(3)$: