The polynomial \(x^{101} + Ax + B\) is divisible by \(x^2 + x + 1\) for some real numbers \(A\) and \(B\). Find \(A+B\).
+397 \(\displaystyle \left\{\frac{x^{102}-1}{x^{3}-1}-\frac{x^{101}-x^{2}}{x^{3}-1}\right\}(x^{2}+x+1)=x^{101}+x+1, \\ \text{so} \\ A+B=2.\)
The two fractions inside the curly brackets are both the sums of GP's.
The first is the sum of
\(1+x^{3}+x^{6}+ \dots +x^{99},\)
and the second the sum of
\(x^{2}+x^{5}+ \dots + x^{98}.\)
