Three Times Sum of Cubes of Three Indeterminates Plus 6 Times their Product
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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
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity: Exercise $5 \ \text {(ii)}$