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

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(A) Given that $1+\omega+\omega^2 = 0$,we have $1+\omega^2 = -\omega$ and $1+\omega = -\omega^2$.Substituting these into the expression:$(1-\omega+\om

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$r, r \in N$

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(A) Given that $1+\omega+\omega^2 = 0$,we have $1+\omega^2 = -\omega$ and $1+\omega = -\omega^2$.Substituting these into the expression:$(1-\omega+\omega^2)^{3k} + (1-\omega^2+\omega)^{3k} = (1-\omega+\omega^2)^{3k+1} + (1+\omega-\omega^2)^{3k+1}$$(-2\omega)^{3k} + (-2\omega^2)^{3k} = (-2\omega)^{3k+1} + (-2\omega^2)^{3k+1}$$(-2)^{3k} \omega^{3k} + (-2)^{3k} \omega^{6k} = (-2)^{3k+1} \omega^{3k+1} + (-2)^{3k+1} \omega^{2(3k+1)}$Since $\omega^3 = 1$,$\omega^{3k} = 1$ and $\omega^{6k} = 1$:$(-2)^{3k}(1+1) = (-2)^{3k} \cdot (-2) \cdot (\omega + \omega^2)$$2 = -2(\omega + \omega^2)$Since $\omega + \omega^2 = -1$,we get $2 = -2(-1) = 2$.This identity holds for all $k \in N$. Thus,$k = r$ where $r \in N$.

If $1, \omega, \omega^2$ are the cube roots of unity,$k$ is a positive integer and $(1-\omega+\omega^2)^{3k} + (1-\omega^2+\omega)^{3k} = (1-\omega+\omega^2)^{3k+1} + (1+\omega-\omega^2)^{3k+1}$,then $k=$

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