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

From ProofWiki
Jump to navigation Jump to search

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:

\(\ds \paren {a + b + c}^3\) \(=\) \(\ds \paren {a + b + c} \paren {a^2 + b^2 + c^2 + 2 a b + 2 a c + 2 b c}\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds a^3 + b^3 + c^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 3 b^2 c + 3 a c^2 + 3 b c^2 + 6 a b c\)


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

\(\ds \paren {a + b \omega + c \omega^2}^3\) \(=\) \(\ds a^3 + b^3 \omega^3 + c^3 \omega^6 + 3 a^2 b \omega + 3 a^2 c \omega^2 + 3 a b^2 \omega^2 + 3 b^2 c \omega^4 + 3 a c^2 \omega^4 + 3 b c^2 \omega^5 + 6 a b c \omega^3\)
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds a^3 + b^3 + c^3 + 3 a^2 b \omega + 3 a^2 c \omega^2 + 3 a b^2 \omega^2 + 3 b^2 c \omega + 3 a c^2 \omega + 3 b c^2 \omega^2 + 6 a b c\) as $\omega^3 = 1$


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

\(\ds \paren {a + b \omega^2 + c \omega}^3\) \(=\) \(\ds a^3 + b^3 \omega^6 + c^3 \omega^3 + 3 a^2 b \omega^2 + 3 a^2 c \omega + 3 a b^2 \omega^4 + 3 b^2 c \omega^5 + 3 a c^2 \omega^2 + 3 b c^2 \omega^4 + 6 a b c \omega^3\)
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds a^3 + b^3 + c^3 + 3 a^2 b \omega^2 + 3 a^2 c \omega + 3 a b^2 \omega + 3 b^2 c \omega^2 + 3 a c^2 \omega^2 + 3 b c^2 \omega + 6 a b c\) as $\omega^3 = 1$


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

\(\ds \) \(\) \(\ds \paren {a + b + c}^3 + \paren {a + b \omega + c \omega^2}^3 + \paren {a + b \omega^2 + c \omega}^3\)
\(\ds \) \(=\) \(\ds a^3 + b^3 + c^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 3 b^2 c + 3 a c^2 + 3 b c^2 + 6 a b c\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds a^3 + b^3 + c^3 + 3 a^2 b \omega + 3 a^2 c \omega^2 + 3 a b^2 \omega^2 + 3 b^2 c \omega + 3 a c^2 \omega + 3 b c^2 \omega^2 + 6 a b c\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds a^3 + b^3 + c^3 + 3 a^2 b \omega^2 + 3 a^2 c \omega + 3 a b^2 \omega + 3 b^2 c \omega^2 + 3 a c^2 \omega^2 + 3 b c^2 \omega + 6 a b c\)
\(\ds \) \(=\) \(\ds 3 \paren {a^3 + b^3 + c^3 + 6 a b c}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {3 a^2 b + 3 a^2 c + 3 a b^2 + 3 b^2 c + 3 a c^2 + 3 b c^2} \paren {1 + \omega + \omega^2}\)
\(\ds \) \(=\) \(\ds 3 \paren {a^3 + b^3 + c^3 + 6 a b c}\) Sum of Cube Roots of Unity: $1 + \omega + \omega^2 = 0$

$\blacksquare$


Sources