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If $\alpha$ and $\beta$ are imaginary cube roots of units then value of $\alpha^4+\beta^{28}+\frac{1}{\alpha \beta}$ is
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Let $\alpha=\omega, \beta=\omega^2$
Now $\alpha^4+\beta^{28}+\frac{1}{\alpha \beta}$
$\begin{aligned} & =\omega^4+\left(\omega^2\right)^{28}+\frac{1}{\omega \omega^2} \\ & =\omega^4+\omega^{56}+\frac{1}{\omega^3} \\ & =\omega+\omega^2+\frac{1}{1} \\ & \left.=0 \quad \quad \quad \text { [as } \omega^3=1\right]\end{aligned}$
Now $\alpha^4+\beta^{28}+\frac{1}{\alpha \beta}$
$\begin{aligned} & =\omega^4+\left(\omega^2\right)^{28}+\frac{1}{\omega \omega^2} \\ & =\omega^4+\omega^{56}+\frac{1}{\omega^3} \\ & =\omega+\omega^2+\frac{1}{1} \\ & \left.=0 \quad \quad \quad \text { [as } \omega^3=1\right]\end{aligned}$
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