Question

If $\alpha$ and $\beta$ are imaginary cube roots of unity, then ${\alpha }^4+{\beta }^4+\frac{1}{\alpha \beta }$
[IIT 1977]

• A. 3
• B. 1
• C. 2

Answer 1

The correct option is A 0
Complex cube root of unity are $1,\omega ,{\omega }^2$
Let $\alpha =\omega ,\beta ={\omega }^2;$ Then ${\alpha }^4+{\beta }^4+{\alpha }^{-1}{\beta }^{-1}$
$={\omega }^4+({\omega }^2{)}^4+({\omega }^{-1}\left)\right({\omega }^2{)}^{-1}=\omega +{\omega }^2+1=0.$