## Factorising the Equation for Roots of Unity

The cube roots of unity are the roots of
$\omega^3 =1 \rightarrow \omega^3 -1=0$
. There are three cube roots of unity. The sixth roots of unity are the roots of
$\omega^6-1=0$
. There are six xis th roots of unity.
All the cube roots of unity are sixth roots of unity. This means that a factor of
$\omega^6-1=0$
is
$\omega^3 -1$
. In fact,
$\omega^6-1=(\omega^3+1)(\omega^3-1)$
.
The ninth roots of unity are the roots of nbsp;
$\omega^99-1=0$
. There are nine ninth roots of unity.
All the cube roots of unity are ninth roots of unity. This means that a factor of
$\omega^9-1=0$
is
$\omega^3 -1$
. In fact,
$\omega^9-1=(\omega^6 +\omega^3+1)(\omega^3-1)$
.
In fact, the
$3nth$
roots of unity are the roots of the equation
$\omega^{3n} -1=0$

All the cube roots of unity are
$3nth$
roots of unity, so as before
$\omega^3 -1$
should be a factor of
$\omega^{3n} -1=0$
. In fact,
$\omega^{3n} -1=(\omega^{3n-3} + \omega^{3n-6} + ...+ \omega^3 + 1)( \omega^3 -1)$

In fact, if If there are m roots of unity, the solutions of the equation
$\omega^{n} -1=0$
and
$m$
is any divisor of
$n$
then
\begin{equation} \begin{aligned} \omega^{n} -1 &= (\omega^{n-m} + \omega^{n-2m} + ...+ \omega^m + 1)( \omega^m -1) \\ &= (\omega^{n-m} + \omega^{n-2m} + ...+ \omega^m + 1)( \omega^{m-1} + \omega^{m-2} + \omega^{m-3} + ...+ \omega + +1)(\omega-1) \end{aligned} \end{equation}