If alpha is a non-real root of x^7=1,then alp | vedclass

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Question

(C) Given that $\alpha$ is a non-real root of $x^7=1$,we have $\alpha^7=1$ and $\alpha 
eq 1$. Expanding the expression: $\alpha(1+\alpha)(1+\alpha^2

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) Given that $\alpha$ is a non-real root of $x^7=1$,we have $\alpha^7=1$ and $\alpha 
eq 1$. Expanding the expression: $\alpha(1+\alpha)(1+\alpha^2+\alpha^4) = \alpha(1+\alpha^2+\alpha^4+\alpha+\alpha^3+\alpha^5)$ $= \alpha + \alpha^3 + \alpha^5 + \alpha^2 + \alpha^4 + \alpha^6$ $= \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6$ This is a geometric series sum: $= \frac{\alpha(1-\alpha^6)}{1-\alpha} = \frac{\alpha-\alpha^7}{1-\alpha}$ Since $\alpha^7=1$,we have: $= \frac{\alpha-1}{1-\alpha} = -1$.

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

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