<s>[INST] How to find $\omega^7$ and $\omega^6$ from $\omega^5+1=0$ 

I did the first parts : 
$$\omega= (\cos \pi + i \sin \pi)^\frac{1}{5} \implies \omega^5 = \cos \pi + i \sin \pi=-1$$

$\omega=-1$ is a root so : 
$$\omega^5-1= (\omega+1) (\omega^4-\omega^3+\omega^2-\omega+1)=0$$
$$\omega^4-\omega^3+\omega^2-\omega=-1$$

$$\omega-\omega^4=\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} - (\cos \frac{4\pi}{5} + i \sin \frac{4\pi}{5})$$
$$\omega-\omega^4=\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} - (-\cos \frac{\pi}{5} + i \sin \frac{\pi}{5})$$
$$\omega-\omega^4= 2 \cos \frac{\pi}{5}$$
Did similar method for $\omega^3-\omega^2$ and got the desired result.

For this part I only know the first part which is: 
$$\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}$$
$$2\cos {\pi}{5} + 2\cos \frac{3\pi}{5}=\omega-\omega^4+\omega^3-\omega^2=1$$
$$\cos \frac{\pi}{5} + \cos \frac{3\pi}{5}=\frac{1}{2}$$

$$\cos \frac{\pi}{5} \cos \frac{3\pi}{5} = \frac{1}{4} (\omega-\omega^4)(\omega^3-\omega^2)$$
$$\frac{1}{4} (\omega^4-\omega^3-\omega^7+\omega^6)$$
How do I find $\omega^7$ and $\omega^6$ ???? please help.

Also I would love if I receive support for the last part too, I don't know how  to make a quadratic equation
 [/INST] For the quadratic equation part, note that if $(x-a)(x-b)=0$ the roots of the equation are $a$ and $b$.
Expand the brackets and you get the equation $x^2-(a+b)x+ab=0$ with roots $a$ and $b$.
This means that if you know the sum of the roots $a+b$ and the product of the roots $ab$, you can write down a quadratic equation which has the roots $a$ and $b$. This is what your question is leading you through.
</s>