The remainder of a huge power

1. The remainder has degree at most $1$: $P(x)=(x-1)(x-2)Q(x)+ax+b$. Evaluating at $1$ and $2$ gives $a+b=2$ and $2a+b=3$, so $a=1$, $b=1$. The remainder is $x+1$.

2. Since $x^3-1$ is divisible by $x^2+x+1$, we have $x^3\equiv1$ modulo $x^2+x+1$. Now $2026=3\cdot675+1$, so
$$x^{2026}=\left(x^3\right)^{675}x\equiv x.$$
Therefore $x^{2026}+x+1\equiv2x+1$. As $2x+1$ has degree $<2$, it is the remainder.

Remark. Equivalently, with $\omega=e^{2\pi i/3}$: $\omega^{2026}+\omega+1=2\omega+1$, and the remainder is the unique polynomial of degree $\le1$ taking the same values at $\omega$ and $\bar\omega$.

Lesson. Work "modulo" the divisor: replace $x^2+x+1$ by $0$ (or $x^3$ by $1$) and reduce the exponents.