Question

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

• A. $1$
• B. $\omega$
• C. ${\omega }^2$
• D. $0$

Answer 1

The correct option is D $0$

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

Now we know that ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$ since '$\omega$' is the cube root of unity.

Hence

${\alpha }^4+{\beta }^4+(\alpha \beta {)}^{-1}$

$=({\alpha }^2+{\beta }^2{)}^2-2{\alpha }^2{\beta }^2+\frac{1}{\alpha \beta }$

$=\left[\right(\alpha +\beta {)}^2-2\alpha \beta {]}^2-2{\alpha }^2{\beta }^2+\frac{1}{\alpha \beta }$

$=\left[\right(\omega +{\omega }^2{)}^2-2{\omega }^3{]}^2-2{\omega }^6+\frac{1}{{\omega }^3}$

$=[1-2{]}^2-2+1$

$=(-1{)}^2-1$

$=0$