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(C) The complex cube roots of unity are $1, \omega, \omega^2$. Since $\alpha$ and $\beta$ are the complex cube roots,we have $\alpha = \omega$ and $\b

Vedclass · Accepted Answer

$3$

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(C) The complex cube roots of unity are $1, \omega, \omega^2$. Since $\alpha$ and $\beta$ are the complex cube roots,we have $\alpha = \omega$ and $\beta = \omega^2$ (or vice versa).Given expression: $\alpha^3 + \beta^3 + \alpha^{-2} 	imes \beta^{-2} = \alpha^3 + \beta^3 + \frac{1}{(\alpha \beta)^2}$.Since $\alpha = \omega$ and $\beta = \omega^2$,we have $\alpha^3 = \omega^3 = 1$ and $\beta^3 = (\omega^2)^3 = \omega^6 = 1$.Also,$\alpha \beta = \omega 	imes \omega^2 = \omega^3 = 1$.Substituting these values: $1 + 1 + \frac{1}{(1)^2} = 1 + 1 + 1 = 3$.

If $\alpha$ and $\beta$ are the complex cube roots of unity,then $\alpha^3+\beta^3+\alpha^{-2} \times \beta^{-2}$ is equal to

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