If $\alpha$ is a non-real root of the equation $x^6-1=0$, then $\frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1}$ is equal to

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Aicharya · Accepted Answer

Since, $\alpha$ is a non-real root of the equation $x^6-1=0$. $\begin{array}{lc}	herefore & \alpha^6-1=0 \ \Rightarrow & \alpha^6-1=0 	imes \alpha-1 \ \Rightarrow & \frac{\alpha^6-1}{\alpha-1}=0 \ \Rightarrow & 1+\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5=0 \ \Rightarrow & (1+\alpha)+\alpha^2(1+\alpha)+\alpha^4(1+\alpha)=0 \ \Rightarrow & (1+\alpha)\left(1+\alpha^2+\alpha^4ight)=0\end{array}$ $\begin{aligned} & 	herefore \frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1}=\frac{\alpha^2(\alpha+1)+\alpha^4(\alpha+1)}{\alpha+1} \ & \quad=\frac{(\alpha+1)\left(\alpha^2+\alpha^4ight)}{\alpha+1} \ & =\alpha^2+\alpha^4=-1 \quad 	ext { [from Eq. (i)] }\end{aligned}$

If $\alpha$ is a non-real root of the equation $x^6-1=0$, then…

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