$\omega$ 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.
Dividing the polynomial $x^{10}+x^5+3$ by $x^2+x+1$ leaves a remainder of $2$. Plugging in $x=\omega$ yields your answer is $2$.
Since $$\omega^3-1=(\omega-1)(\omega^2+\omega+1)=0,$$ we obtain:
$$\omega^{10}+\omega^5+3=\omega^{10}-\omega+\omega^5-\omega^2+\omega^2+\omega+3=$$
$$=\omega(\omega^9-1)+\omega^2(\omega^3-1)+(\omega^2+\omega+1)+2=2.$$