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(D) The roots of the equation $x^2+x+1=0$ are the complex cube roots of unity, $\omega$ and $\omega^2$.Let $\alpha = \omega$ and $\beta = \omega^2$.We

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$x^2+x+1=0$

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(D) The roots of the equation $x^2+x+1=0$ are the complex cube roots of unity, $\omega$ and $\omega^2$.Let $\alpha = \omega$ and $\beta = \omega^2$.We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$, which implies $\omega + \omega^2 = -1$.Calculating the new roots:$\alpha^{19} = \omega^{19} = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$.$\beta^7 = (\omega^2)^7 = \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.The required quadratic equation with roots $\alpha^{19}$ and $\beta^7$ is $x^2 - (\alpha^{19} + \beta^7)x + (\alpha^{19} \cdot \beta^7) = 0$.Substituting the values:$x^2 - (\omega + \omega^2)x + (\omega \cdot \omega^2) = 0$.Since $\omega + \omega^2 = -1$ and $\omega^3 = 1$, the equation becomes:$x^2 - (-1)x + 1 = 0$, which simplifies to $x^2 + x + 1 = 0$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2+x+1=0$, then the equation whose roots are $\alpha^{19}$ and $\beta^7$ is

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