If $\alpha$ is a non-real root of $x^7=1$, then $\alpha(1+\alpha)\left(1+\alpha^2+\alpha^4\right)=$

If $\alpha$ is a non-real root of $x^7=1$, then $\alpha(1+\alpha)\left(1+\alpha^2+\alpha^4\right)=$
  1. 1
  2. 2
  3. -1
  4. -2

Solution

We have given that $\alpha$ is non-real root of $x^7=1$ $ \therefore \alpha^7=1 \text { and } \alpha \neq 1 $ Now, $\alpha(1+\alpha)\left(1+\alpha^2+\alpha^4\right)$ $ \begin{aligned} & =\alpha\left(1+\alpha^2+\alpha^4+\alpha+\alpha^3+\alpha^5\right) \\ & =\alpha+\alpha^3+\alpha^5+\alpha^2+\alpha^4+\alpha^6 \\ & =\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6 \\ & =\frac{\alpha\left(1-\alpha^6\right)}{1-\alpha} \quad[\because \alpha \neq 1] \\ & =\frac{\alpha-\alpha^7}{1-\alpha}=\frac{\alpha-1}{1-\alpha} \quad\left[\because \alpha^7=1\right] \\ & \therefore \quad \alpha(1+\alpha)\left(1+\alpha^2+\alpha^4\right)=-1 \end{aligned} $

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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