What is the remainder when $(x + 1)^{2010}$ is divided by $x^2 + x + 1$?
I know that when I divide any polynomial by a quadratic, the remainder will have a degree less than the quadratic, so it will be at most linear. That means I can write the remainder as $ax + b$ for some constants $a$ and $b$. To find $a$ and $b$, I need to use the fact that the remainder and the dividend have the same value at any root of the divisor. The roots of $x^2 + x + 1$ are the complex numbers $\frac{-1 \pm i \sqrt{3}}{2}$, which are also the cube roots of unity. So, let me plug in these values into $(x + 1)^{2010}$ and $ax + b$ and set them equal. For the first root, I get $(\frac{-1 + i \sqrt{3}}{2} + 1)^{2010} = a (\frac{-1 + i \sqrt{3}}{2}) + b$. I can simplify the left side by noting that $\frac{-1 + i \sqrt{3}}{2} + 1 = \frac{1 + i \sqrt{3}}{2}$, which is one of the sixth roots of unity. So, raising it to the power of 2010, which is a multiple of 6, gives me 1. Therefore, I have $1 = a (\frac{-1 + i \sqrt{3}}{2}) + b$. Similarly, for the second root, I get $(\frac{-1 - i \sqrt{3}}{2} + 1)^{2010} = a (\frac{-1 - i \sqrt{3}}{2}) + b$. Again, I can simplify the left side by noting that $\frac{-1 - i \sqrt{3}}{2} + 1 = \frac{1 - i \sqrt{3}}{2}$, which is another sixth root of unity. So, raising it to the power of 2010 also gives me 1. Therefore, I have $1 = a (\frac{-1 - i \sqrt{3}}{2}) + b$. Now I have two equations with two unknowns, $a$ and $b$. I can solve them by elimination. If I add the two equations, I get $2 = a (\frac{-1 + i \sqrt{3}}{2} + \frac{-1 - i \sqrt{3}}{2}) + 2b$. Simplifying, I get $2 = -a + 2b$. If I subtract the second equation from the first, I get $0 = a (\frac{-1 + i \sqrt{3}}{2} - \frac{-1 - i \sqrt{3}}{2}) + 0b$. Simplifying, I get $0 = a i \sqrt{3}$. This implies that $a = 0$. Plugging this into the first equation, I get $1 = b$. So, the remainder is $0x + 1$, or just 1. 