If 1, omega, omega^2 denote the cube roots of | vedclass

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(D) 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

Vedclass · Accepted Answer

$32$

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(D) We know that $1+\omega+\omega^2=0$,which implies $1+\omega^2=-\omega$ and $1+\omega=-\omega^2$. Substituting these into the expression: $(1-\omega+\omega^2)^5+(1+\omega-\omega^2)^5 = (-\omega-\omega)^5+(-\omega^2-\omega^2)^5$ $= (-2\omega)^5+(-2\omega^2)^5$ $= -32\omega^5 - 32\omega^{10}$ $= -32(\omega^5+\omega^{10})$ Since $\omega^3=1$,we have $\omega^5 = \omega^2$ and $\omega^{10} = \omega$. $= -32(\omega^2+\omega)$ Since $1+\omega+\omega^2=0$,we have $\omega^2+\omega = -1$. $= -32(-1) = 32$.

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

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