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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

wlad
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Rohini Sharma
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    Start by expanding the expression as a polynomial in $\omega$, then reducing it accordingly as $\omega^3 = 1$. – hardmath Jun 28 '15 at 16:37
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    Try multiplying out term by term to get various powers of $\omega$. Some may cancel (be careful about signs) and others can be reduced. Since $\omega$ is a root of $x^3-1=0$ and $x\neq 1$ you should be able to find a quadratic satisfied by $\omega$ - if you did that first you might see a simpler way to get the answer. – Mark Bennet Jun 28 '15 at 16:39

5 Answers5

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

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

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

robjohn
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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 $4\omega^3=4$

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

mich95
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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 $P = 3-(-1) = 4$.

DeepSea
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