The given problem asks for the limit of a complex expression. Here's the breakdown:

We are given the condition:

$\lim_{x \to 2} \left( \sqrt[3]{\frac{a}{x^3} + \frac{b}{x^3} - 1} \right) = A$

Now, we are asked to find:

$\lim_{x \to 2} \left( \sqrt[3]{\frac{a}{27} + \frac{b}{27x^3} + x - 1} \right)$

Step 1: Calculate the first limit

For the first limit, we need to analyze the expression when $x \to 2$. Substituting $x = 2$:

$\lim_{x \to 2} \left( \sqrt[3]{\frac{a}{x^3} + \frac{b}{x^3} - 1} \right)$

At $x = 2$, we get:

$\frac{a}{2^3} + \frac{b}{2^3} - 1 = \frac{a}{8} + \frac{b}{8} - 1 = \frac{a + b}{8} - 1$

This equals $A$, so:

$A = \frac{a + b}{8} - 1 \quad \Rightarrow \quad A = \frac{a + b - 8}{8}$

Step 2: Calculate the second limit

Now, for the second expression, substitute $x = 2$ into:

$\lim_{x \to 2} \left( \sqrt[3]{\frac{a}{27} + \frac{b}{27x^3} + x - 1} \right)$

Substituting $x = 2$:

[ \frac{a}{27} + \frac{b}{27(2^3)} + 2 - 1 = \frac{a}{27} + \frac{b}{27 \times 8} + 1 = \frac{a}{27} + \frac{b}{216} + 1 \