Question

If $\alpha$ is a non-real root of $x^6=1$, then $\frac{{\alpha }^5+{\alpha }^3+\alpha +1}{{\alpha }^2+1}$ is equal to

• A. ${\alpha }^2$
• B. 0
• C. $-{\alpha }^2$
• D. $\alpha$

Answer 1

Answer: (C)

Given that
$x^6=1$
$\Rightarrow x^6-1=0...(i)$
$\Rightarrow (x-1)\left(x^5+x^4+x^3+x^2+x+1\right)=0$
$\to x^5+x^4+x^3+x^2+x+1=0$
$[\because$ roots are non-real]
Since $\alpha$ is a root of the equation (i)
$\therefore {\alpha }^5+{\alpha }^4+{\alpha }^3+{\alpha }^2+\alpha +1=0$
$\Rightarrow {\alpha }^5+{\alpha }^3+\alpha +1=-\left({\alpha }^4+{\alpha }^2\right)$
$\Rightarrow {\alpha }^5+{\alpha }^3+\alpha +1=-{\alpha }^2\left({\alpha }^2+1\right)$
$\frac{{\alpha }^5+{\alpha }^3+\alpha +1}{{\alpha }^2+1}=-{\alpha }^2$