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

Question

Dr. Abubacker · Accepted Answer

(b)
The complex cube roots of unity are $1,\ \omega,\ \omega^{2}$
Let $\alpha = \omega,\ \ \ \beta = \omega^{2}$
Then, $\alpha^{4} + \beta^{4} + \alpha^{- 1}\beta^{- 1} = \omega^{4} + \left( \omega^{2} \right)^{4} + (\omega)^{- 1}\left( \omega^{2} \right)^{- 1}$
$= \omega + \omega^{2} + 1 = 0$

Q. If $\alpha$ and $\beta$ are imaginary cube roots of unity, then $\alpha^{4} + \beta^{4} + \frac{1}{\alpha\beta}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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