Question

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

• A. 1
• B. -1
• C. 0
• D. None of these

Answer 1

Answer: (C)

Since $\alpha$ and $\beta$ are complex roots of unity,
we may write $\alpha =\omega ,\beta ={\omega }^2$
Hence, ${\alpha }^4+{\beta }^{28}+\frac{1}{\alpha \beta }={\omega }^4+{\left({\omega }^2\right)}^{28}+\frac{1}{\omega \cdot {\omega }^2}$
$=\omega +{\omega }^{56}+1=\omega +{\omega }^2+1=0$