We are tasked with rationalizing the denominator of the expression

[ \frac{2}{\sqrt[3]{4} + \sqrt[3]{32}}. ]

Step 1: Simplify the cube roots

First, simplify the cube roots in the denominator. Recall that:

[ \sqrt[3]{4} = 4^{1/3} \quad \text{and} \quad \sqrt[3]{32} = 32^{1/3}. ]

We can rewrite $32 = 2^5$, so:

[ \sqrt[3]{32} = 2^{5/3} = 2 \cdot 2^{2/3} = 2 \sqrt[3]{4}. ]

Thus, the expression becomes:

[ \frac{2}{\sqrt[3]{4} + 2 \sqrt[3]{4}} = \frac{2}{3 \sqrt[3]{4}}. ]

Step 2: Rationalize the denominator

Now, to rationalize the denominator $3 \sqrt[3]{4}$, multiply both the numerator and the denominator by $\sqrt[3]{16}$, which is the cube root of $4^2$. This will eliminate the cube root in the denominator:

[ \frac{2}{3 \sqrt[3]{4}} \times \frac{\sqrt[3]{16}}{\sqrt[3]{16}} = \frac{2 \sqrt[3]{16}}{3 \sqrt[3]{64}}. ]

Since $\sqrt[3]{64} = 4$, the expression simplifies to:

[ \frac{2 \sqrt[3]{16}}{3 \times 4} = \frac{\sqrt[3]{16}}{6}. ]

Step 3: Final form and computation of $A + B$

The expression is now in the form $\frac{\sqrt[3]{A}}{B}$, where $A = 16$ and $B = 6$.

Thus, the minimum value of $A + B$ is:

$A + B = 16 + 6 = 22.$

Final Answer:

The minimum possible value of $A + B$ is $\boxed{22}$.

Do you want more details or have any questions?

Here are some related questions for further practice:

1. Rationalize $\frac{1}{\sqrt[3]{5} + \sqrt[3]{25}}$.
2. Simplify the expression $\sqrt[3]{8} + \sqrt[3]{16}$.
3. Find the cube root of $64 \times 8$ and express it in simplest terms.
4. If $x = \sqrt[3]{2}$, express $x^2 + 3x + 1$ in simplest form.
5. Rationalize the denominator of $\frac{4}{\sqrt[3]{7} - \sqrt[3]{49}}$.

Tip: When rationalizing cube roots, multiply by a power of the cube root that will complete the cube.