If omega is a cube root of unity,then the val | vedclass

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(B) Given that $\omega$ is a cube root of unity,we know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.Using $1 + \omega^2 = -\omega$,the first t

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$32$

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(B) Given that $\omega$ is a cube root of unity,we know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.Using $1 + \omega^2 = -\omega$,the first term becomes $(1 - \omega + \omega^2)^5 = (-\omega - \omega)^5 = (-2\omega)^5 = -32\omega^5$.Since $\omega^5 = \omega^3 \cdot \omega^2 = \omega^2$,this simplifies to $-32\omega^2$.Using $1 + \omega = -\omega^2$,the second term becomes $(1 + \omega - \omega^2)^5 = (-\omega^2 - \omega^2)^5 = (-2\omega^2)^5 = -32\omega^{10}$.Since $\omega^{10} = (\omega^3)^3 \cdot \omega = \omega$,this simplifies to $-32\omega$.Adding the two terms: $-32\omega^2 - 32\omega = -32(\omega^2 + \omega)$.Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$.Therefore,$-32(-1) = 32$.

If $\omega$ is a cube root of unity,then the value of $(1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5$ is:

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