If omega is a non-real root of the equation x | vedclass

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(B) We know that $1 + \omega^r + \omega^{2r}$ is a geometric series sum.If $r$ is a multiple of $3$,then $\omega^r = 1$ and $\omega^{2r} = 1$,so $1 +

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$3$

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(B) We know that $1 + \omega^r + \omega^{2r}$ is a geometric series sum.If $r$ is a multiple of $3$,then $\omega^r = 1$ and $\omega^{2r} = 1$,so $1 + \omega^r + \omega^{2r} = 1 + 1 + 1 = 3$.If $r$ is not a multiple of $3$,then $1 + \omega^r + \omega^{2r} = \frac{1 - (\omega^r)^3}{1 - \omega^r} = \frac{1 - (\omega^3)^r}{1 - \omega^r} = \frac{1 - 1}{1 - \omega^r} = 0$.For $r = 1, 2, 3, 4, 5$,the terms are:$r=1: 1 + \omega + \omega^2 = 0$$r=2: 1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$$r=3: 1 + \omega^3 + \omega^6 = 1 + 1 + 1 = 3$$r=4: 1 + \omega^4 + \omega^8 = 1 + \omega + \omega^2 = 0$$r=5: 1 + \omega^5 + \omega^{10} = 1 + \omega^2 + \omega = 0$Sum $= 0 + 0 + 3 + 0 + 0 = 3$.

If $\omega$ is a non-real root of the equation $x^3 - 1 = 0$,then the value of $\sum_{r=1}^{5} (1 + \omega^r + \omega^{2r})$ is

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