# If n is a natural number that is not a multiple of 3, prove that 1+w^n+w^(2n)=0?

Aug 7, 2018

Please go through the Explanation.

#### Explanation:

${w}^{3 n} - 1 = {\left({w}^{3}\right)}^{n} - 1 - 1 = 0. \ldots \ldots \ldots \ldots . \left({\ast}^{1}\right)$.

Factorising, ${w}^{3 n} - 1 = {\left({w}^{n}\right)}^{3} - {1}^{3} ,$

$= \left({w}^{n} - 1\right) \left\{{\left({w}^{n}\right)}^{2} + {w}^{n} \cdot 1 + {1}^{2}\right\}$,

$\therefore {w}^{3 n} - 1 = \left({w}^{n} - 1\right) \left({w}^{2 n} + {w}^{n} + 1\right) \ldots \ldots \left({\ast}^{2}\right)$.

Combining $\left({\ast}^{1}\right) \mathmr{and} \left({\ast}^{2}\right)$, we get,

$\left({w}^{n} - 1\right) \left({w}^{2 n} + {w}^{n} + 1\right) = 0$

$\Rightarrow {w}^{n} - 1 = 0 , \mathmr{and} , {w}^{2 n} + {w}^{n} + 1 = 0$.

Since, $n \ne 3 m , m \in \mathbb{N} , {w}^{n} \ne 1$.

$\therefore {w}^{2 n} + {w}^{n} + 1 = 0 , \mathmr{if} , n \ne 3 m , m \in \mathbb{N}$.