\(\lim\limits_{x\to\infty}\left(\frac{Ax^3}{x^2 + x + 1} - B(x + 1)\right) = 8\)
\(= \lim\limits_{x\to \infty} \frac{Ax^3 - Bx^3 - 2Bx^2 - 2Bx - B}{x^2+ x + 1}=8\)
\(= \lim\limits_{x\to\infty}\cfrac{x^2 \left[(A - B)x - 2B - \frac{2B}x - \frac B{x^2}\right]}{x^2\left(1 + \frac1 x + \frac1 {x^2}\right)} = 8\)
Limit exists if A - B = 0
⇒ A = B
Then -2B = 8
⇒ \(B = -4\)
\(∴ A = -4\)
\(\therefore (A +B) = |-4 -4| = 8\)