Given Equation is $x^{2}+x+1=0$
$\Rightarrow x=\frac{-1 \pm \sqrt{3} i}{2}$
Thus, roots are $\omega$ and $\omega^{2}$ $\alpha=\omega$ and $\beta=\omega^{2}$
$\mathrm{S} o, \alpha^{19}=(\omega)^{19}=\left(\omega^{3}\right)^{6}, \omega=\omega \quad\left(\because \omega^{3}=1\right)$
and $\beta^{7}=\left(\omega^{2}\right)^{7}=\omega^{14}=\left(\omega^{3}\right)^{4} \cdot \omega^{2}=\omega^{2}$
Now, $\alpha^{19}+\beta^{7}=\omega+\omega^{2}=-1\left(\because 1+\omega+\omega^{2}=0\right)$
and $\left(\alpha^{19}\right)\left(\beta^{7}\right)=(\omega)\left(\omega^{2}\right)=1$
$\therefore$ Required Quadratic equation whose roots are $\alpha^{19}$
and $\beta^{7}$ is
$x^{2}+\left(\alpha^{19}+\beta^{7}\right) x+\left(\alpha^{19}\right)\left(\beta^{7}\right)=0$
$\Rightarrow x^{2}+(-1) x+1=0$
$\Rightarrow x^{2}-x+1=0$