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(B) Let $\alpha = \omega$ and $\beta = \omega^2$,where $\omega$ is an imaginary cube root of unity.We know that $\omega^3 = 1$ and $1 + \omega + \omeg

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$0$

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(B) Let $\alpha = \omega$ and $\beta = \omega^2$,where $\omega$ is an imaginary cube root of unity.We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Now,substitute these into the expression $\alpha^4 + \beta^{28} + \frac{1}{\alpha \beta}$:$= \omega^4 + (\omega^2)^{28} + \frac{1}{\omega \cdot \omega^2}$$= \omega^4 + \omega^{56} + \frac{1}{\omega^3}$Since $\omega^3 = 1$,we have $\omega^4 = \omega^3 \cdot \omega = \omega$ and $\omega^{56} = (\omega^3)^{18} \cdot \omega^2 = 1^{18} \cdot \omega^2 = \omega^2$.Thus,the expression becomes $\omega + \omega^2 + \frac{1}{1} = \omega + \omega^2 + 1$.Since $1 + \omega + \omega^2 = 0$,the value is $0$.

If $\alpha$ and $\beta$ are imaginary cube roots of unity,then the value of $\alpha^4 + \beta^{28} + \frac{1}{\alpha \beta}$ is

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