Question

If $\alpha$, $\beta$ are the complex cube roots of unity then ${\alpha }^{100}+{\beta }^{100}+\frac{1}{{\alpha }^{100}\times {\beta }^{100}}=$

• A. $-1$
• B. $1$
• C. $\alpha$
• D. $0$

Answer 1

The correct option is D $0$

Let $\alpha =w$ and $\beta =w^2$
Then
${\alpha }^{100}=w^{100}$
$=w^{99}.w$
$=w$ ..(i)
And
${\beta }^{100}$
$=(w^2{)}^{100}$
$=w^{200}$
$=w^{198}.w^2$
$=w^2$ ...(ii)
Now
$(\alpha \beta {)}^{100}$
$=(w\times w^2{)}^{100}$
$=(w^3{)}^{100}$
$=1$...(iii)
Hence
${\alpha }^{100}+{\beta }^{100}+\frac{1}{(\alpha \beta {)}^{100}}$
$=w^2+w+1$
$=0$