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(C) We know that $\omega$ is a cube root of unity,so $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Since $n$ is not a multiple of $3$,$n$ can be of t

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(C) We know that $\omega$ is a cube root of unity,so $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Since $n$ is not a multiple of $3$,$n$ can be of the form $3k + 1$ or $3k + 2$ for some integer $k \ge 0$.Case $1$: If $n = 3k + 1$,then $\omega^n = \omega^{3k+1} = (\omega^3)^k \cdot \omega = 1^k \cdot \omega = \omega$ and $\omega^{2n} = \omega^{6k+2} = (\omega^3)^{2k} \cdot \omega^2 = 1^{2k} \cdot \omega^2 = \omega^2$.Thus,$1 + \omega^n + \omega^{2n} = 1 + \omega + \omega^2 = 0$.Case $2$: If $n = 3k + 2$,then $\omega^n = \omega^{3k+2} = (\omega^3)^k \cdot \omega^2 = \omega^2$ and $\omega^{2n} = \omega^{6k+4} = (\omega^3)^{2k+1} \cdot \omega = \omega$.Thus,$1 + \omega^n + \omega^{2n} = 1 + \omega^2 + \omega = 0$.In both cases,the sum is $0$.

If $n$ is a positive integer not a multiple of $3$,then $1 + \omega^n + \omega^{2n} = $

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