If $a$ is a root of $x^2+ x + 1$, simplify $1 + a + a^2 +\dots+ a^{2017}.$

I haven't looked through your working, but here's a much easier method:

If $a^2+a+1=0$, then $(a-1)(a^2+a+1)=0$, that is, $a^3-1=0$, ie, $a^3=1$.

On the other hand, $1+a+\dots+a^{2017}=\frac{a^{2018}-1}{a-1}$. Since $a^3=1$, it follows that $a^{2018}=a^2$. So, your sum simplifies to $\frac{a^2}{a-1}$. <-- wrong! Read on....

Edit:

I had a typo (thanks @Math lover for pointing that out). The sum simplifies to $\frac{a^2-1}{a-1}$, which simplifies easily to $a+1$.