The value of (1 - omega + omega^2)(1 - omega^ | vedclass

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(C) We know that $1 + \omega + \omega^2 = 0$. Therefore,$1 + \omega^2 = -\omega$ and $1 + \omega = -\omega^2$. Substituting these into the expression:

Vedclass · Accepted Answer

$-128\omega$

Vedclass · Answer

(C) We know that $1 + \omega + \omega^2 = 0$. Therefore,$1 + \omega^2 = -\omega$ and $1 + \omega = -\omega^2$. Substituting these into the expression: $(1 - \omega + \omega^2)(1 - \omega^2 + \omega)^6 = (-\omega - \omega)(-\omega^2 - \omega^2)^6$ $= (-2\omega)(-2\omega^2)^6$ $= (-2\omega)(64\omega^{12})$ Since $\omega^3 = 1$,$\omega^{12} = (\omega^3)^4 = 1^4 = 1$. $= (-2\omega)(64 	imes 1) = -128\omega$.

The value of $(1 - \omega + \omega^2)(1 - \omega^2 + \omega)^6$,where $\omega, \omega^2$ are the complex cube roots of unity.

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