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 $
$\rightarrow x^{5}+x^{4}+x^{3}+x^{2}+x+1=0$
$[\because$ 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}$