# omega is a cube root of unity. Show the following?

## $\left(a + b + c\right) \left(a + b \omega + c {\omega}^{2}\right) \left(a + b {\omega}^{2} + c \omega\right) = {a}^{3} + {b}^{3} + {c}^{3} - 3 a b c$

Jul 20, 2017

We are given that $\omega$ is a cube root of unity; therefore $\omega$ satisfies the equation

${z}^{3} = 1$

$\therefore {z}^{3} - 1 = 0$
$\therefore \left(z - 1\right) \left({z}^{2} + z + 1\right) = 0$

From which can get the actual values of the roots:

$z = 1 , \frac{- 1 \pm \sqrt{3} i}{2}$

Thus some easily verifiable properties are that:

• The roots can be denoted by $1 , \omega , {\omega}^{2}$ where $\omega$ is any of the complex roots.
• $1 + \omega + {\omega}^{2} = 0 \implies \omega + {\omega}^{2} = - 1$.
• ${\omega}^{3} = 1 \implies {\omega}^{4} = \omega$. etc

So now, let us consider the RHS of the given expression:

$R H S = \left(a + b + c\right) \left(a + b \omega + c {\omega}^{2}\right) \left(a + b {\omega}^{2} + c \omega\right)$

$\text{ } = \left(a + b + c\right) \left({a}^{2} + a b {\omega}^{2} + a c \omega + a b \omega + {b}^{2} {\omega}^{3} + b c {\omega}^{2} + a c {\omega}^{2} + b c {\omega}^{4} + {c}^{2} {\omega}^{3}\right)$

 " " = (a+b+c) (a^2+b^2+c^2+acomega+abomega+bcomega+abomega^2+bcomega^2+ acomega^2+

$\text{ } = \left(a + b + c\right) \left({a}^{2} + {b}^{2} + {c}^{2} + a c \left(\omega + {\omega}^{2}\right) + a b \left(\omega + {\omega}^{2}\right) + b c \left(\omega + {\omega}^{2}\right)\right)$

$\text{ } = \left(a + b + c\right) \left({a}^{2} + {b}^{2} + {c}^{2} - a c - a b - b c\right)$

$\text{ } = {a}^{3} + a {b}^{2} + a {c}^{2} - {a}^{2} c - {a}^{2} b - a b c + {a}^{2} b + {b}^{3} + b {c}^{2} - a b c - a {b}^{2} - {b}^{2} c + {a}^{2} c + {b}^{2} c + {c}^{3} - a {c}^{2} - a b c - b {c}^{2}$

And the almost all cross terms cancel, leaving:

$R H S = {a}^{3} + {b}^{3} + {c}^{3} - 3 a b c \setminus \setminus \setminus$ QED