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(B) Given the equation $x^2-x+1=0$. The roots of this equation are $\omega$ and $\omega^2$,where $\omega$ is the complex cube root of unity. Let $\alp

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$x^2+x+1=0$

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(B) Given the equation $x^2-x+1=0$. The roots of this equation are $\omega$ and $\omega^2$,where $\omega$ is the complex cube root of unity. Let $\alpha = \omega$ and $\beta = \omega^2$. We need to find the equation with roots $\alpha^{2015}$ and $\beta^{2015}$. $\alpha^{2015} = \omega^{2015} = \omega^{3 	imes 671 + 2} = \omega^2$. $\beta^{2015} = (\omega^2)^{2015} = \omega^{4030} = \omega^{3 	imes 1343 + 1} = \omega$. The sum of the new roots is $\alpha^{2015} + \beta^{2015} = \omega^2 + \omega = -1$. The product of the new roots is $\alpha^{2015} \cdot \beta^{2015} = \omega^2 \cdot \omega = \omega^3 = 1$. The required quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$. Substituting the values,we get $x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.

If $\alpha, \beta$ are the roots of $x^2-x+1=0$,then the quadratic equation whose roots are $\alpha^{2015}$ and $\beta^{2015}$ is

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