Find the value of (1 + 2omega + omega^2)^{3n} | vedclass

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(A) We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega^2 = -\omega$ and $1 + \omega = -\omega^2$.Substituting these into the expressio

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(A) We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega^2 = -\omega$ and $1 + \omega = -\omega^2$.Substituting these into the expression:$(1 + \omega^2 + 2\omega)^{3n} - (1 + \omega + 2\omega^2)^{3n} = (-\omega + 2\omega)^{3n} - (-\omega^2 + 2\omega^2)^{3n}$$= (\omega)^{3n} - (\omega^2)^{3n}$$= (\omega^3)^n - (\omega^3)^{2n}$Since $\omega^3 = 1$,we have $1^n - 1^{2n} = 1 - 1 = 0$.

Find the value of $(1 + 2\omega + \omega^2)^{3n} - (1 + \omega + 2\omega^2)^{3n}$.

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