We are tasked to calculate $A = \frac{2}{\sqrt[3]{9} + \sqrt[3]{3} + 1}$ and find $(A + 1)^3$.

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Step 1: Simplify $A$

Let: $a = \sqrt[3]{3}.$ Then: $A = \frac{2}{a^2 + a + 1}.$

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Step 2: Add 1 to $A$

$A + 1 = \frac{2}{a^2 + a + 1} + 1 = \frac{2 + (a^2 + a + 1)}{a^2 + a + 1}.$ Simplify the numerator: $A + 1 = \frac{a^2 + a + 3}{a^2 + a + 1}.$

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Step 3: Cube $A + 1$

We need to compute: $(A + 1)^3 = \left( \frac{a^2 + a + 3}{a^2 + a + 1} \right)^3.$ Let $x = a^2 + a + 1$. Then: $A + 1 = \frac{x + 2}{x}.$ Cubing it: $(A + 1)^3 = \left( \frac{x + 2}{x} \right)^3 = \frac{(x + 2)^3}{x^3}.$ Expand $(x + 2)^3$: $(x + 2)^3 = x^3 + 3x^2(2) + 3x(2^2) + 2^3 = x^3 + 6x^2 + 12x + 8.$ Thus: $(A + 1)^3 = \frac{x^3 + 6x^2 + 12x + 8}{x^3}.$ Simplify: $(A + 1)^3 = 1 + \frac{6x^2}{x^3} + \frac{12x}{x^3} + \frac{8}{x^3}.$ $(A + 1)^3 = 1 + \frac{6}{x} + \frac{12}{x^2} + \frac{8}{x^3}.$

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Step 4: Substitute back $x = a^2 + a + 1$

Since $a = \sqrt[3]{3}$, $x = a^2 + a + 1 = 1 + \sqrt[3]{3} + \sqrt[3]{9}$. Substituting this back to simplify the cube requires detailed rationalization or numerical approximation.

Would you like further simplification or a numerical evaluation?