1. **Problem statement:** Find $a+b$ given that $$\lim_{x \to 2} \frac{(x^3 - 8)(x^2 + ax + b)}{(x^2 - 4)^2} = 6.$$ 2. **Rewrite and analyze the limit:** Note that $x^3 - 8 = (x-2)(x^2 + 2x + 4)$ and $x^2 - 4 = (x-2)(x+2)$. So the denominator is $((x-2)(x+2))^2 = (x-2)^2 (x+2)^2$. 3. **Substitute factorizations:** $$\lim_{x \to 2} \frac{(x-2)(x^2 + 2x + 4)(x^2 + ax + b)}{(x-2)^2 (x+2)^2} = \lim_{x \to 2} \frac{(x^2 + 2x + 4)(x^2 + ax + b)}{(x-2)(x+2)^2}.$$ 4. **Since the limit exists and is finite, the factor $(x-2)$ in the denominator must be canceled by the numerator.** This means $x^2 + ax + b$ must be zero at $x=2$: $$4 + 2a + b = 0 \implies b = -4 - 2a.$$ 5. **Rewrite the limit using this:** $$\lim_{x \to 2} \frac{(x^2 + 2x + 4)(x^2 + ax + b)}{(x-2)(x+2)^2} = \lim_{x \to 2} \frac{(x^2 + 2x + 4)(x^2 + ax + b)}{(x-2)(x+2)^2}.$$ Since $x^2 + ax + b$ has root at $x=2$, factor it as $(x-2)(x + c)$ for some $c$: $$x^2 + ax + b = (x-2)(x+c) = x^2 + (c-2)x - 2c.$$ Matching coefficients: $$a = c - 2, \quad b = -2c.$$ 6. **Substitute back into the limit:** $$\lim_{x \to 2} \frac{(x^2 + 2x + 4)(x-2)(x+c)}{(x-2)(x+2)^2} = \lim_{x \to 2} \frac{(x^2 + 2x + 4)(x+c)}{(x+2)^2}.$$ Cancel $(x-2)$. 7. **Evaluate the limit by direct substitution:** $$\frac{(2^2 + 2\cdot 2 + 4)(2 + c)}{(2 + 2)^2} = \frac{(4 + 4 + 4)(2 + c)}{4^2} = \frac{12(2 + c)}{16} = \frac{3}{4}(2 + c).$$ 8. **Set equal to 6 and solve for $c$:** $$\frac{3}{4}(2 + c) = 6 \implies 2 + c = 8 \implies c = 6.$$ 9. **Find $a$ and $b$:** $$a = c - 2 = 6 - 2 = 4,$$ $$b = -2c = -12.$$ 10. **Calculate $a + b$:** $$a + b = 4 + (-12) = -8.$$ **Final answer:** $\boxed{-8}$ corresponds to option 2.