If alpha is a non-real root of the equation x | vedclass

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(D) The given equation is $x^6-1=0$,which implies $x^6=1$. Since $\alpha$ is a root,$\alpha^6=1$. Also,$\alpha 
eq 1$ because $\alpha$ is a non-real

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) The given equation is $x^6-1=0$,which implies $x^6=1$. Since $\alpha$ is a root,$\alpha^6=1$. Also,$\alpha 
eq 1$ because $\alpha$ is a non-real root (the roots of $x^6-1=0$ are $e^{i2k\pi/6}$ for $k=0, 1, 2, 3, 4, 5$,and only $k=0$ gives $x=1$). We can write the numerator as $\alpha^2(1+\alpha+\alpha^2+\alpha^3)$. Since $\alpha^6-1 = (\alpha-1)(\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1) = 0$ and $\alpha 
eq 1$,we have $\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1 = 0$. This implies $\alpha^5+\alpha^4+\alpha^3+\alpha^2 = -\alpha-1 = -(\alpha+1)$. Substituting this into the expression: $\frac{-(\alpha+1)}{\alpha+1} = -1$.

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} = $

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