algebra precalculus - $omega$ is a solution of $x^2+x+1=0$, find $omega^{10}+omega^5+3$

I am working on a scholarship exam practice assuming high school or pre-university math knowledge. I am stuck at the question below:

Let $\omega$ be a solution of the equation $x^2+x+1=0$. Then $\omega^{10}+\omega^5+3=.....$

My first question is how it would be possible since the discriminant of $x^2+x+1=0$ is less than $0$ so I am not sure how I can continue or start from here. The answer key provided is $2$. Please advise.

Answer

$\omega$ is such that $\omega^2+\omega+1=0$ i.e. $\omega^2=-\omega-1$

Therefore $\omega^3=\omega \cdot \omega^2=\omega(-\omega-1)=-\omega^2-\omega=1$

$\omega^5=\omega^3 \cdot \omega^2=\omega^2=-\omega-1$

$\omega^{10}=(\omega^5)^2=(-\omega-1)^2=\omega^2+1+2\omega=\omega$

Hence $\omega^{10}+\omega^5+3=\omega-\omega-1+3=2$