+0

0
80
1

If $$\omega^3 = 1$$ and $$\omega \neq 1$$, then compute $$(1 - \omega + \omega^2)(1 + \omega - \omega^2)$$.

I've tried multiplying the equation out and then replacing all the $$\omega^3$$ with 1 but then I just end up with $$1-\omega-\omega^2$$ and I don't know what to do from there. Thanks!

Jan 17, 2021

#1
+25640
+1

If $$\omega^3 = 1$$ and $$\omega \neq 1$$,

then compute $$(1 - \omega + \omega^2)(1 + \omega - \omega^2)$$.

My attempt:

$$\begin{array}{|rcll|} \hline \mathbf{ (1 - \omega + \omega^2)(1 + \omega - \omega^2) } \\ &=& \Big(1 - (\omega - \omega^2)\Big) \Big(1 + (\omega - \omega^2)\Big) \\ &=& 1 - (\omega - \omega^2)^2 \\ &=& 1 - (\omega^2 - 2\omega^3+\omega^4) \quad | \quad \omega^3=1 \\ &=& 1 - (\omega^2 - 2+\omega^4) \quad | \quad \omega^4=\omega^3*\omega=\omega \\ &=& 1 - (\omega^2 - 2+\omega) \\ &=& 3 - (\omega^2 + \omega) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (\omega^2 + \omega)^2 &=& \omega^4 + 2\omega^3 + \omega^2 \quad | \quad \omega^3=1 \\ (\omega^2 + \omega)^2 &=& \omega^4 + 2 + \omega^2 \quad | \quad \omega^4=\omega^3*\omega=\omega \\ (\omega^2 + \omega)^2 &=& \omega + 2 + \omega^2 \\ (\omega^2 + \omega)^2 &=& \omega^2 + \omega + 2 \\ (\omega^2 + \omega)^2- (\omega^2 + \omega) - 2 &=& 0 \quad | \quad \omega^2 + \omega = x \\ \mathbf{x^2-x-2} &=& \mathbf{0} \\\\ x &=& \dfrac{1\pm\sqrt{1-4*(-2)}}{2} \\ x &=& \dfrac{1\pm\sqrt{9}}{2} \\ x &=& \dfrac{1\pm3}{2} \\\\ x_1 &=& \dfrac{1+3}{2} \\ x_1&=& 2 \\ \omega^2 + \omega &=& 2 \qquad \text{no solution!} \\ && \boxed{\omega \ne 1 } \\\\ x_2 &=& \dfrac{1-3}{2} \\ x_2 &=& -1 \\ \mathbf{\omega^2 + \omega }&=& \mathbf{-1} \\ \mathbf{ (1 - \omega + \omega^2)(1 + \omega - \omega^2) } &=& 3 - (\omega^2 + \omega) \\ &=& 3 - (-1) \\ \mathbf{ (1 - \omega + \omega^2)(1 + \omega - \omega^2) }&=& \mathbf{ 4 } \\ \hline \end{array}$$

Jan 17, 2021