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

Question

Aicharya · Accepted Answer

Given that

$\begin{aligned} & \Rightarrow(x-1)\left(x^5+x^4+x^3+x^2+x+1\right)=0 \\ & \Rightarrow \quad x^5+x^4+x^3+x^2+x+1=0 \\ & \end{aligned}$ $[\because$ roots are non-real] Since $\alpha$ is a root of the equation (i) $\begin{array}{ll} \therefore & \alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0 \\ \Rightarrow & \alpha^5+\alpha^3+\alpha+1=-\left(\alpha^4+\alpha^2\right) \\ \Rightarrow & \alpha^5+\alpha^3+\alpha+1=-\alpha^2\left(\alpha^2+1\right) \\ & \frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=-\alpha^2 \end{array}$

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

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