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(C) Given $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$. Since the roots of $x^2+x+1=0$ are the imaginary cube roots of unity,we have $\al

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(C) Given $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$. Since the roots of $x^2+x+1=0$ are the imaginary cube roots of unity,we have $\alpha = \omega$ and $\beta = \omega^2$,where $\omega^3 = 1$ and $1+\omega+\omega^2=0$. Thus,$\alpha^3 = 1$ and $\beta^3 = 1$. Also,$\alpha+\beta = -1$ and $\alpha\beta = 1$. We need to evaluate the expression $E = (2-\alpha)(2-\beta)(2-\alpha^{10})(2-\alpha^{20})$. Since $\alpha^3 = 1$,we have $\alpha^{10} = (\alpha^3)^3 \cdot \alpha = \alpha$ and $\alpha^{20} = (\alpha^3)^6 \cdot \alpha^2 = \alpha^2$. Substituting these into the expression: $E = (2-\alpha)(2-\beta)(2-\alpha)(2-\alpha^2)$. Since $\beta = \alpha^2$ (as $\alpha\beta=1$ and $\alpha^3=1$),we have: $E = (2-\alpha)(2-\alpha^2)(2-\alpha)(2-\alpha^2) = [(2-\alpha)(2-\alpha^2)]^2$. Expanding the inner term: $(2-\alpha)(2-\alpha^2) = 4 - 2\alpha^2 - 2\alpha + \alpha^3 = 4 - 2(\alpha+\alpha^2) + 1$. Since $1+\alpha+\alpha^2=0$,we have $\alpha+\alpha^2 = -1$. So,$(2-\alpha)(2-\alpha^2) = 4 - 2(-1) + 1 = 4+2+1 = 7$. Therefore,$E = 7^2 = 49$.

If $\alpha, \beta$ are the roots of the equation $1+x+x^2=0$,then $(2-\alpha)(2-\beta)(2-\alpha^{10})(2-\alpha^{20})=$

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