Large powers can produce small results

$8x^2-10x+3 =0 \quad \Rightarrow (2x-1)(4x-3) = 0 \\ \Rightarrow \alpha = \frac{1}{2} \quad \quad \beta^2 = \frac{3}{4} \quad \Rightarrow \beta = \frac{\sqrt{3}}{2}$

We note that:

$(\alpha + i\beta)^{100} = \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^{100} = \left( e^{i\frac{\pi}{3}}\right)^{100} = e^{i\frac{100\pi}{3}} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \\ (\alpha - i\beta)^{100} = e^{-i\frac{100\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$

The equation, whose roots are $(\alpha + i\beta)^{100}$ and $(\alpha - i \beta ) ^{100}$ is:

$x^2 - \left((\alpha + i\beta)^{100} + (\alpha - i\beta)^{100} \right) x + (\alpha + i\beta)^{100}(\alpha - i\beta)^{100} = 0 \\ x^2 - \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2} -\frac{1}{2} + i\frac{\sqrt{3}}{2} \right) x + \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) \left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} \right) = 0 \\ \Rightarrow \boxed {x^2+x+1=0}$

$8x^2-10x+3=0\Rightarrow \alpha =\frac{1}{2},\beta=\frac{\sqrt{3}}{2}$ $\left(\alpha+i\beta\right)^{100}=\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)^{100}\\=\left(\cos(60^\circ)+\sin(60^\circ)\right)^{100}=\cos(6000^\circ)+\sin(6000^\circ)i\mbox{[By De Moivre's formula]}\\=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$

Similarly, we obtain

$\left(\alpha-i\beta\right)^{100}=\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)^{100}\\=\left(\cos(300^\circ)+\sin(300^\circ)\right)^{100}=\cos(30000^\circ)+\sin(30000^\circ)i\mbox{[By De Moivre's formula]}\\=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$

Let the final equation be of the form $x^2+bx+c$ . By Vieta's theorem, we have

$b=-(\left(\alpha-i\beta\right)^{100}+\left(\alpha+i\beta\right)^{100})=-\left(-\frac{1}{2}-\frac{1}{2}\right)=1$

Similarly we have

$c=(\left(\alpha-i\beta\right)^{100}\left(\alpha+i\beta\right)^{100})=\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=\frac{1}{4}+\frac{3}{4}=1$

We thus arrive at $x^2+x+1=0$ .