Rationalize the denominator of $\frac{2}{\sqrt[3]{4}+\sqrt[3]{32}}$. The answer can be written in the form of $\frac{\sqrt[3]{A}}{B}$, where $A$ and $B$ are positive integers. Find the minimum possible value of $A+B$.

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.

Rationalize the denominator of $\frac{2}{\sqrt[3]{4} + \sqrt[3]{32}}$. The answer can be written in the form of $\frac{\sqrt[3]{A}}{B}$, where $A$ and $B$ are positive integers. Find the minimum possible value of $A+B$.

Learn how to rationalize the denominator of $\frac{2}{\sqrt[3]{4} + \sqrt[3]{32}}$. This problem explores cube roots, algebraic rationalization, and the steps to simplify expressions involving cube roots. Suitable for students in Grades 9-12.

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