Question

If $\alpha$ and $\beta$ are the complex cube-roots of unity, then ${\alpha }^4+{\beta }^4+{\alpha }^{-1}{\beta }^{-1}=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

Cube roots of unity are $1$, $\omega$ and ${\omega }^2$

where $\omega$ and ${\omega }^2$ are non-real complex cube roots of unity.

Now, the properties of cube roots of unity are:

${1+\omega +\omega }^2=0$ and ${\omega }^3=1$

If $\alpha =\omega$ and $\beta ={\omega }^2$

then

${\alpha }^4+{\beta }^4+{\alpha }^{-1}{\beta }^{-1}={\omega }^4+{\left({\omega }^2\right)}^4+\frac{1}{\omega \cdot {\omega }^2}$

$={\omega }^4+{\omega }^8+\frac{1}{{\omega }^3}$

$={\omega }^3\cdot \omega +{\left({\omega }^3\right)}^2\cdot {\omega }^2+1$

$=\omega +{\omega }^2+1$

$=0$