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

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$x^2+x+1=0$

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(D) For the quadratic equation $x^2+x+1=0$,the roots are the complex cube roots of unity,$\omega$ and $\omega^2$.Thus,$\alpha = \omega$ and $\beta = \omega^2$.We need to find the equation with roots $\alpha^{2021}$ and $\beta^{2021}$.Since $\omega^3 = 1$,we have $\alpha^{2021} = \omega^{2021} = (\omega^3)^{673} \cdot \omega^2 = \omega^2$.Similarly,$\beta^{2021} = (\omega^2)^{2021} = \omega^{4042} = (\omega^3)^{1347} \cdot \omega = \omega$.The new roots are $\omega^2$ and $\omega$.The quadratic equation with roots $\omega$ and $\omega^2$ is given by $x^2 - (\omega + \omega^2)x + \omega \cdot \omega^2 = 0$.Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$ and $\omega^3 = 1$.Substituting these values,we get $x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.

List-$I$	List-$II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$	$1.$ True
$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\|1-z_1\|\|1-z_2\| \ldots \|1-z_9\|}{10}$ equals	$3.$ $1$
$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals	$4.$ $2$

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2+x+1=0$,then the equation whose roots are $\alpha^{2021}$ and $\beta^{2021}$ is given by $.......$

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