If alpha is a root of the equation x^2-x+1=0, | vedclass

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(D) Given the equation $x^2-x+1=0$. Since $\alpha$ is a root,$\alpha^2-\alpha+1=0$. Dividing by $\alpha$,we get $\alpha-1+\frac{1}{\alpha}=0$,so $\alp

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$-9$

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(D) Given the equation $x^2-x+1=0$. Since $\alpha$ is a root,$\alpha^2-\alpha+1=0$. Dividing by $\alpha$,we get $\alpha-1+\frac{1}{\alpha}=0$,so $\alpha+\frac{1}{\alpha}=1$. Now,$\left(\alpha+\frac{1}{\alpha}\right)^2 = \alpha^2+\frac{1}{\alpha^2}+2 = 1^2 = 1$,which implies $\alpha^2+\frac{1}{\alpha^2} = -1$. Also,$\alpha^3+\frac{1}{\alpha^3} = (\alpha+\frac{1}{\alpha})(\alpha^2+\frac{1}{\alpha^2}-1) = (1)(-1-1) = -2$. For $\alpha^4+\frac{1}{\alpha^4}$,we use $(\alpha^2+\frac{1}{\alpha^2})^2 = \alpha^4+\frac{1}{\alpha^4}+2 = (-1)^2 = 1$,so $\alpha^4+\frac{1}{\alpha^4} = -1$. Substituting these values into the expression: $(1)^3 + (-1)^3 + (-2)^3 + (-1)^3 = 1 - 1 - 8 - 1 = -9$.

If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3=$

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