1. ## demostration polynomial

Hl! I have problems with demostration

Let $\displaystyle \zeta=e^{2\pi i/n}$ be a primitive nth root unity

For numbers $\displaystyle a$ and $\displaystyle b$, prove that
$\displaystyle a^n-b^n= (a-b) (a-\zeta b)(a- {\zeta}^2b) \cdot{} \cdot{} \cdot{} (a-\zeta^{n-1}b)$

and if $\displaystyle n$ is odd , that
$\displaystyle a^n+b^n= (a+b) (a+\zeta b)(a+ {\zeta}^2b) \cdot{} \cdot{} \cdot{} (a+\zeta^{n-1}b)$

Hint: Set $\displaystyle x=a/b$ if $\displaystyle b\neq{0}$

I do not know how to start or how to use the hint

2. ## Re: demostration polynomial

$a^n - b^n = b^n \left(\left(\dfrac a b\right)^n - 1\right)$

$x = \dfrac a b,~b \neq 0$

$a^n - b^n = b^n (x^n - 1)$

Consider the polynomial $x^n-1$

It's roots are precisely the $n$ roots of unity and further if $\zeta = e^{2\pi i \frac \theta n}$ then these roots are given by $\zeta^k,~k \in 0,~1,~2,\dots n-1$

thus we can write

$x^n - 1 = \displaystyle \prod_{k=1}^{n-1} \left(x-\zeta^k\right)$

$a^n - b^n = b^n \displaystyle \prod_{k=1}^{n-1} \left(x-\zeta^k\right)$

and you should be able to take it from here.

3. ## Re: demostration polynomial

There's a little more to it than that. If $\displaystyle \zeta = \frac{2\pi i}{n}$ where $\displaystyle i$ divides $\displaystyle n$, it is not a generator for the full set of roots.

4. ## Re: demostration polynomial

Originally Posted by Archie
There's a little more to it than that. If $\displaystyle \zeta = \frac{2\pi i}{n}$ where $\displaystyle i$ divides $\displaystyle n$, it is not a generator for the full set of roots.
I should have explicitly noted this but I believe the fact that $\zeta$ is stated to be a primitive root overcomes this.

5. ## Re: demostration polynomial

$\displaystyle e^{2\pi i/n}$ is a primitive n-th root of unity

Here $\displaystyle i=\sqrt{-1}$

6. ## Re: demostration polynomial

thanks to all