Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ where $\omega^3 = 1$ If $\omega^3 = 1$ and $\omega \neq 1$, then compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$
I'm pretty lost, I don't really know where to start.
Thanks
 A: $$(1 - \omega + \omega^2)(1 + \omega - \omega^2)$$
$$=((1 + \omega^2) - \omega)(1 + \omega - \omega^2)$$
$$=(-\omega - \omega)(-\omega^2 - \omega^2)$$
$$=(-2\omega)(-2\omega^2)$$
$$=4\omega^3=4$$
Where $1+\omega+\omega^2=0$ and $\omega^3=1$
A: $$
\begin{align}
(1-\omega+\omega^2)(1+\omega-\omega^2)
&=1-\omega^2+2\omega^3-\omega^4\\
&=1-\omega^2+2-\omega\\
&=4-(1+\omega+\omega^2)\\
&=4
\end{align}
$$
The last step is true since $(1+\omega+\omega^2)\overbrace{(1-\omega)}^{\text{not $0$}}=1-\omega^3=0$.
A: Note that $x^3-1=(x-1)(x^2+x+1)$ and $\omega\neq 1$ so that $\omega^2+\omega+1=0$
This means that $1+\omega-\omega^2=-2\omega^2$ and $1-\omega+\omega^2=-2\omega$
The product then reduces to $4\omega^3=4$
A: $(1-(\omega-\omega^{2}))(1+\omega-\omega^{2})=1-(\omega-\omega^{2})^{2}=1-(\omega^{4}+\omega^{2}-2\omega^{3})=1-(\omega+\omega^{2}-2)=3-(\omega+\omega^{2})=3-(-1)=4$
A: Let $P$ be the product, then: $P = 1^2-(\omega - \omega^2)^2=1-(\omega^2-2\omega^3+\omega^4)=1-\omega^2+2-\omega^4=3-\omega^2-\omega$. Observe that $1-\omega^3= (1-\omega)(1+\omega+\omega^2) = 0$ since $1 \neq \omega$. So: $-\omega - \omega^2 = 1$, and $P = 3-(-1) = 4$.
