7. $[\mathbf{6}]$ What is the remainder when $(1+x)^{2010}$ is divided by $1+x+x^{2}$ ?
Answer: 1 We use polynomial congruence $\bmod 1+x+x^{2}$ to find the desired remainder. Since $x^{2}+x+1 \mid x^{3}-1$, we have that $x^{3} \equiv 1\left(\bmod 1+x+x^{2}\right)$. Now:
$$
\begin{aligned}
(1+x)^{2010} & \equiv\left(-x^{2}\right)^{2010} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv x^{4020} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv\left(x^{3}\right)^{1340} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv 1^{1340} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv 1 \quad\left(\bmod 1+x+x^{2}\right)
\end{aligned}
$$

Thus, the answer is 1 .