A Complex Remainder Problem

Since $x^2 + x + 1$ is a factor of $f(x)$ , its zeroes, $\omega$ and $\omega^2$ (see note below), must also be roots of $f(x)$ . Then we have

$\begin{array}{rllll} f(\omega) = \ \ 0 \\ \\ \omega^{2018} + A\ \omega + B - 2018 = \ \ 0 \\ \\ (\omega^3)^{672} \cdot \omega^2 +A\ \omega + B - 2018 = \ \ 0 \\ \\ \omega^2 +A\ \omega + B - 2018 = \ \ 0 & & & \small \text{[ Since } \omega^3 = 1 \ ] \\ \\ (\omega^2 + \omega + 1) + (A - 1)\ \omega + B - 2019 \ \ = 0 \\ \\ (A - 1)\ \omega + B - 2019 \ \ = 0 & & & \small \text{[ Since } \omega^2 + \omega + 1 = 0 \ ] \end{array}$

Now we just need to equate coefficients. Since $\omega$ is complex, we must have $A - 1 = 0$ and $B - 2019 = 0$ so our answer is $A + B = \boxed{2020}$

Note: The equation $x^3 = 1$ can be re-written as $(x - 1)(x^2 + x + 1) = 0$ ; the second factor yields the complex roots $\frac{-1 \pm \sqrt{3}i}{2}$ . These complex cube roots of unity are usually called $\omega$ and $\omega^2$ as it's easily verified that each is the square of the other.

$f(x)=x^{2018}+Ax+B-2018$

We can let $f(x)=(x^2+x+1)Q(x)$

$x^{2018}+Ax+B-2018=(x^2+x+1)Q(x)$

$\Rightarrow x^{2018}-2018=(x^2+x+1)Q(x)-Ax-B$

We know that $x^3-1=(x-1)(x^2+x+1)$

So we can rewrite $x^{2018}-2018$ as follows:

$x^{2018}-2018$

$=(x^{2018}-x^{2015})+(x^{2015}-x^{2012})+(x^{2012}-x^{2009})+...+(x^{5}-x^{2})+x^2-2018$

$=x^{2015}(x^3-1)+x^{2012}(x^3-1)+x^{2009}(x^3-1)+...+x^{2}(x^3-1)+x^2-2018$

$=(x^3-1)(x^{2015}+x^{2012}+x^{2009}+...+x^{2})+x^2-2018$

$=(x^2+x+1)(x-1)(x^{2015}+x^{2012}+x^{2009}+...+x^{2})+x^2-2018$

$=(x^2+x+1)(x-1)(x^{2015}+x^{2012}+x^{2009}+...+x^{2})+x^2+x+1-x-2019$

$=(x^2+x+1)[(x-1)(x^{2015}+x^{2012}+x^{2009}+...+x^{2})+1]-x-2019$

By comparing coeffiecients, $A=1, B=2019$

$\therefore A+B=2020$

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Alternative approach: Complex numbers-Euler's formula

Let $f(x)=(x^2+x+1)Q(x)$

When $x^2+x+1=0$

$x=\cfrac{-1\pm\sqrt{3}i}{2}=e^{\pm\cfrac{2\pi}{3}}$

By Remainder Factor Theorem ,

$\begin{cases} f(e^{\cfrac{2\pi}{3}})=0\Rightarrow e^{\cfrac{2\pi\cdot 2018}{3}}+Ae^{\cfrac{2\pi}{3}}+B-2018=0\Rightarrow e^{\cfrac{4\pi}{3}}+Ae^{\cfrac{2\pi}{3}}+B-2018=0 \\ f(e^{-\cfrac{2\pi}{3}})=0\Rightarrow e^{\cfrac{-2\pi\cdot 2018}{3}}+Ae^{\cfrac{-2\pi}{3}}+B-2018=0\Rightarrow e^{\cfrac{-4\pi}{3}}+Ae^{\cfrac{-2\pi}{3}}+B-2018=0 \end{cases}$

$\Rightarrow \begin{cases} cos\cfrac{4\pi}{3}+isin\cfrac{4\pi}{3}+A(cos\cfrac{2\pi}{3}+isin\cfrac{2\pi}{3})+B-2018=0 \\ cos\cfrac{-4\pi}{3}+isin\cfrac{-4\pi}{3}+A(cos\cfrac{-2\pi}{3}+isin\cfrac{-2\pi}{3})+B-2018=0 \end{cases}$

$\Rightarrow \begin{cases} \cfrac{-1}{2}-i\cfrac{\sqrt{3}}{2}+A(\cfrac{-1}{2}+i\cfrac{\sqrt{3}}{2})+B-2018=0\Rightarrow (-\cfrac{1}{2}-\cfrac{1}{2}A+B-2018)+i(-\cfrac{\sqrt{3}}{2}+\cfrac{\sqrt{3}}{2}A)=0 \\ \cfrac{-1}{2}+i\cfrac{\sqrt{3}}{2}+A(\cfrac{-1}{2}+i\cfrac{-\sqrt{3}}{2})+B-2018=0\Rightarrow (-\cfrac{1}{2}-\cfrac{1}{2}A+B-2018)+i(\cfrac{\sqrt{3}}{2}-\cfrac{\sqrt{3}}{2}A)=0 \end{cases}$

By comparing coefficients,

$\begin{cases} -\cfrac{1}{2}-\cfrac{1}{2}A+B-2018=0 \\ \cfrac{\sqrt{3}}{2}-\cfrac{\sqrt{3}}{2}A=0 \end{cases}$

$A=1, -\cfrac{1}{2}-\cfrac{1}{2}+B-2018=0\Rightarrow B-2018-1=0\Rightarrow B=2019$

$\therefore A+B=2020$