I am studying the $n$th roots of unity $1$ and I have a question about this exercise which is showed in my book:
If $\omega$ is one of the imaginary roots of the equation $x^3=1$ ,then find the product $(1-\omega +\omega^2 )\cdot(1+ \omega-\omega^2)$.
Solution: Writing the given equation as $x^3-1=0$,we factor and find $(x-1)(x^2+x+1)=0$.
Since $\omega$ is imaginary,it is a root of $x^2+x+1$ ,so $\omega^2 +\omega +1 =0$.
Hence we have $1+\omega=-\omega^2$ and $1+\omega^2=-\omega$ and $$(1-\omega+\omega^2)(1+\omega -\omega^2)=(-\omega -\omega)(-\omega^2-\omega^2)=4\omega^3=4,$$ since $\omega^3 =1$
Now,my question is about the last step where we have $\omega^3=1$ .If we solve this we have $\omega=1$ ,but previously in my book it's showed that for any root $\omega \ne 1$ we have $\omega^{n-1}+\omega^{n-2}+...+1=0$ as a direct consequence of the fact that $$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+...+1)=0$$
So why do we have now that $\omega=1$ works ?
This leads me to the conclusion that $\omega$ ,being an imaginary root, is such that $\omega=1$ and $\omega \ne 1$ since we have that $\omega^{n-1}+\omega^{n-2}+...+1=0$ if and only if $\omega \ne 1$ and also in the previous exercise we have $\omega^3=1$.
I know I must be wrong but I can't see where's the fall in the logic (I guess I can't draw the conclusion $\omega^3=1 \implies \omega=1 $ but why ?).
I am perplexed and confused at the same time...
