Question

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

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

Answer 1

Answer: (A)

-${\alpha }^2$

$\alpha$ is non real root of $x^6=1$

one possible complex value of

$\alpha =\cos \left(2\pi /6\right)+i\sin \left(2\pi /6\right)$

$\alpha =\frac{1}{2}+\frac{\sqrt{3}}{2}i$

${\alpha }^2=(\frac{1}{2}+\frac{\sqrt{3}}{2}i)(\frac{1}{2}+\frac{\sqrt{3}}{2}i)=\frac{1}{4}+\frac{\sqrt{3}}{4}i+\frac{\sqrt{3}}{4}i-\frac{3}{4}=\frac{-1}{2}+\frac{\sqrt{3}}{2}i$

${\alpha }^3=(\frac{-1}{2}+\frac{\sqrt{3}}{2}i)(\frac{1}{2}+\frac{\sqrt{3}}{2}i)=\frac{-1}{4}=\frac{\sqrt{3}}{4}i+\frac{\sqrt{3}}{4}-\frac{3}{4}=-1$

$\frac{{\alpha }^5+{\alpha }^3+\alpha +1}{{\alpha }^2+1}={\alpha }^3+\frac{\alpha +1}{{\alpha }^2+1}$

$=-1+\frac{(\frac{1}{2}+\frac{\sqrt{3}}{2}i+1)}{(-\frac{1}{2}+\frac{\sqrt{3}}{2}i+1)}$

$=\frac{\frac{1}{2}-\frac{\sqrt{3}}{2}i-1+\frac{1}{2}+\frac{\sqrt{3}}{2}i+1}{(\frac{-1}{2}+\frac{\sqrt{3}}{2}i+1)}$

$=\frac{1}{{\alpha }^2+1}=\frac{{\alpha }^2}{{\alpha }^4+{\alpha }^2}$

$={\alpha }^4={\alpha }^2.{\alpha }^2=(\frac{\sqrt{3}}{2}i-\frac{1}{2})(\frac{\sqrt{3}}{2}i-\frac{1}{2})=-\frac{\sqrt{3}}{2}i-\frac{1}{2}$

$=\frac{{\alpha }^2}{{\alpha }^4+{\alpha }^2}=\frac{{\alpha }^2}{(\frac{-1}{2}+\frac{\sqrt{3}}{2}i-\frac{1}{2}-\frac{\sqrt{3}}{2}i)}=-{\alpha }^2$

Considering option C as $-{\alpha }^2$ , it is correct