But they aren't the same

Note first that $(x^{2} + x + 1)(x - 1) = x^{3} - 1$ , so $\alpha$ and $\beta$ are the cubed roots of unity other than $1.$

By Vieta's we know that $\alpha\beta = 1$ , so whichever root we assign as $\alpha$ and which as $\beta$ we have that $(\alpha\beta)^{7} = 1.$ Also, both of these roots raised to the power of $12$ equal $1$ , as they are both cubed roots of unity. So however we assign $\alpha$ and $\beta$ we have that

$(\alpha)^{19}(\beta)^{7} = (\alpha\beta)^{7}(\alpha)^{12} = \boxed{1}.$

The roots of the equation $x^2 + x + 1 = 0$ are $\alpha$ = $e^{i2\pi/3}$ = cos 2 $\pi$ /3 + i sin 2 $\pi$ /3 and $\beta$ = $e^{i4\pi/3}$ , and the product of the roots of the equation whose roots are $\alpha^{19}$ and $\beta^{7}$ is $\alpha^{19}$ $\beta^{7}$ = $e^{i38\pi/3}$ $e^{i28\pi/3}$ = $e^{i22\pi}$ = 1, Note : I remember $e^{i\alpha}$ = cos $\alpha$ + i sin $\alpha$

W= omega and w^2 are roots of the given eqn . And they ask for w^19 multiplied with (w^2)^7 . Giving w^33 . Using w^3=1 we finally get the answer 1

The roots of $x^{2}+x+1$ are imaginary because $\frac{-1 \pm \sqrt{1-4}}{2}$ = $\frac{-1 \pm \sqrt{3}i}{2}$ . Then we multiply it 19 times, and 7 times, then we should multiply $(\frac{-1 + \sqrt{3}i}{2})^{19}$ and $(\frac{-1 - \sqrt{3}i}{2})^{7}$ . Then $(\frac{-1 + \sqrt{3}i}{2})^{19} \times (\frac{-1 - \sqrt{3}i}{2})^7$ . So we get $1$ .

x = a or x = b x-a = 0 or x-b = 0 (x-a)(x-b) = 0 by multiplying x^2 - (a+b)x + a b = 0 therefore we have our equation x^2 + x + 1 =0 by comparing with the GENERAL equation !!! a b = 1 and as we 1 to the nth power is 1 therefore a^19 * b^7 =1