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

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(D) We know that $1+\omega+\omega^2=0$ and $\omega^3=1$. From $1+\omega+\omega^2=0$,we have $1+\omega^2=-\omega$ and $1+\omega=-\omega^2$. Substitutin

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(D) We know that $1+\omega+\omega^2=0$ and $\omega^3=1$. From $1+\omega+\omega^2=0$,we have $1+\omega^2=-\omega$ and $1+\omega=-\omega^2$. Substituting these into the expression: $\left(1-\omega+\omega^2\right)^6+\left(1-\omega^2+\omega\right)^6 = \left(-\omega-\omega\right)^6+\left(-\omega^2-\omega^2\right)^6$ $= (-2\omega)^6 + (-2\omega^2)^6$ $= 64\omega^6 + 64\omega^{12}$ Since $\omega^3=1$,then $\omega^6 = (\omega^3)^2 = 1^2 = 1$ and $\omega^{12} = (\omega^3)^4 = 1^4 = 1$. $= 64(1) + 64(1) = 64 + 64 = 128$.

If $\omega$ is a complex cube root of unity,then $\left(1-\omega+\omega^2\right)^6+\left(1-\omega^2+\omega\right)^6=$

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