If the roots of $x^2+x-1$ are $\alpha$ and $\beta$, find an $eq^{n}$ whose roots are $\alpha^{19}$ and $\beta^{7}$

If the roots of $x^2+x-1$ are $\alpha$ and $\beta$, find an $eq^{n}$ whose roots are $\alpha^{19}$ and $\beta^{7}$

My Procedure

The roots are $$\frac{-b+\sqrt{b^{2}-4ac}}{2a}$$ and $$\frac{-b-\sqrt{b^{2}-4ac}}{2a}$$ $\therefore$ the roots are $\frac{\sqrt{5}-1}{2}$ and $\frac{-\sqrt{5}-1}{2}$

But this on further solution doesn't give a simple answer. I get $$x^{2}-x-\frac{(\sqrt{5}-1)^{12}}{2^{5}}$$ Is there any other simpler way?

since $$x = 1 - x^2$$ we have $$x^2 = (x^2-1)^2 = x^4 - 2x^2 + 1$$ i.e. $$x^4=3x^2 -1$$ going one step further $$x^8 = (3x^2-1)^2 = 9x^4 - 6x^2 +1 = 21 x^2 -8$$ after one more step you may calculate, e.g., $x^{19}$ as $x^{16}x^2x$
in fact the roots of the original equation are $\frac1{\phi}$ and $-\phi$ where $\phi$ is the golden section.
the question as posed is ambiguous since there are two ways of assigning these values to the $\alpha$ and $\beta$. but in either case if you compute as above you can then do as Dr Graubner suggests and form the equation $(x-\alpha^{19})(x-\beta^7)$, or, equivalently, directly use the standard form for a quadratic in terms of the sum and product of the roots