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(C) Given $n = 3r + 1$ for some integer $r$. We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$. Thus,$1+i\sqrt{3}

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$-(-2)^n$

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(C) Given $n = 3r + 1$ for some integer $r$. We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$. Thus,$1+i\sqrt{3} = -2\omega^2$ and $1-i\sqrt{3} = -2\omega$. Substituting these into the expression: $(1+i\sqrt{3})^n + (1-i\sqrt{3})^n = (-2\omega^2)^n + (-2\omega)^n$ $= (-2)^n (\omega^{2n} + \omega^n)$ Since $n = 3r+1$,$\omega^n = \omega^{3r+1} = \omega$ and $\omega^{2n} = \omega^{6r+2} = \omega^2$. Therefore,the expression becomes $(-2)^n (\omega^2 + \omega)$. Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$. Thus,the result is $(-2)^n (-1) = -(-2)^n$.

If $n$ is an integer which leaves a remainder of $1$ when divided by $3$,then $(1+\sqrt{3}i)^n + (1-\sqrt{3}i)^n$ equals

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