Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ where $\omega^3 = 1$

Question

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

Answer 1

$$(1 - \omega + \omega^2)(1 + \omega - \omega^2)$$ $$=((1 + \omega^2) - \omega)(1 + \omega - \omega^2)$$ $$=(-\omega - \omega)(-\omega^2 - \omega^2)$$ $$=(-2\omega)(-2\omega^2)$$ $$=4\omega^3=4$$

Where $1+\omega+\omega^2=0$ and $\omega^3=1$

Answer 2

$$ \begin{align} (1-\omega+\omega^2)(1+\omega-\omega^2) &=1-\omega^2+2\omega^3-\omega^4\\ &=1-\omega^2+2-\omega\\ &=4-(1+\omega+\omega^2)\\ &=4 \end{align} $$ The last step is true since $(1+\omega+\omega^2)\overbrace{(1-\omega)}^{\text{not $0$}}=1-\omega^3=0$.

Answer 3

$(1-(\omega-\omega^{2}))(1+\omega-\omega^{2})=1-(\omega-\omega^{2})^{2}=1-(\omega^{4}+\omega^{2}-2\omega^{3})=1-(\omega+\omega^{2}-2)=3-(\omega+\omega^{2})=3-(-1)=4$

Answer 4

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 $P = 3-(-1) = 4$.

Answer 5

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 $4\omega^3=4$