If α and β are the roots of the quadratic equation x 2 + x + 1 = 0 , then the equation whose roots…

For quadratic equation, $x^2+x+1=0$

$\Rightarrow x=\frac{-1\pm \sqrt{1-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

$\Rightarrow \mathrm{\alpha }=\frac{-1+\sqrt{3}\mathrm{i}}{2}, \mathrm{\beta }=\frac{-1-\sqrt{3}\mathrm{i}}{2}$

By using cube roots of unity, we can say $\alpha =\omega \& \beta ={\omega }^2$

So, ${\alpha }^{2021}={\omega }^{2021}={\left({\omega }^3\right)}^{673}.{\omega }^2={\omega }^2$, as ${\omega }^3=1$

${\beta }^{2021}={\left({\omega }^2\right)}^{2021}={\left({\omega }^3\right)}^{1347}.\omega =\omega$

Therefore, roots of new equation will be $\omega , {\omega }^2$,

Therefore, equation is $x^2-\left(\omega +{\omega }^2\right)x+\omega .{\omega }^2=0$

We know, $1+\omega +{\omega }^2=0 \& {\omega }^3=1$

$\Rightarrow x^2+x+1=0$ is the required equation.