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

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(D) Given that $\alpha$ is a non-real root of $x^6-1=0$.Since $x^6-1 = (x-1)(x^5+x^4+x^3+x^2+x+1) = 0$,and $\alpha 
eq 1$ (as it is a non-real root),

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$-1$

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(D) Given that $\alpha$ is a non-real root of $x^6-1=0$.Since $x^6-1 = (x-1)(x^5+x^4+x^3+x^2+x+1) = 0$,and $\alpha 
eq 1$ (as it is a non-real root),we have:$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1 = 0$  --- $(i)$We need to evaluate $\frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1}$.Factorizing the numerator:$\alpha^2+\alpha^3+\alpha^4+\alpha^5 = \alpha^2(1+\alpha) + \alpha^4(1+\alpha) = (1+\alpha)(\alpha^2+\alpha^4)$Therefore,the expression becomes:$\frac{(1+\alpha)(\alpha^2+\alpha^4)}{\alpha+1} = \alpha^2+\alpha^4$From equation $(i)$,$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1 = 0$.This can be written as $(1+\alpha) + \alpha^2(1+\alpha) + \alpha^4(1+\alpha) = 0$.$(1+\alpha)(1+\alpha^2+\alpha^4) = 0$.Since $\alpha 
eq -1$ (because $(-1)^6-1 
eq 0$),we must have $1+\alpha^2+\alpha^4 = 0$.Thus,$\alpha^2+\alpha^4 = -1$.

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

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