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The given expression is:

\[
\frac{3a^{-\frac{2}{3}} + 2a^{\frac{1}{3}}}{a^{-\frac{5}{3}}}
\]

Let's break it down and simplify the expression step by step.

### Step 1: Simplify the numerator
The numerator consists of two terms: \( 3a^{-\frac{2}{3}} \) and \( 2a^{\frac{1}{3}} \). It remains as:

\[
3a^{-\frac{2}{3}} + 2a^{\frac{1}{3}}
\]

### Step 2: Simplify the denominator
The denominator is \( a^{-\frac{5}{3}} \).

### Step 3: Apply the exponent rule
Use the property of exponents that states:

\[
\frac{x^m}{x^n} = x^{m-n}
\]

We can rewrite the whole fraction as:

\[
\left( \frac{3a^{-\frac{2}{3}} + 2a^{\frac{1}{3}}}{a^{-\frac{5}{3}}} \right) = (3a^{-\frac{2}{3}} + 2a^{\frac{1}{3}}) \cdot a^{\frac{5}{3}}
\]

### Step 4: Distribute the \( a^{\frac{5}{3}} \)
We distribute \( a^{\frac{5}{3}} \) to each term in the numerator:

1. \( 3a^{-\frac{2}{3}} \cdot a^{\frac{5}{3}} = 3a^{\left(-\frac{2}{3} + \frac{5}{3}\right)} = 3a^{\frac{3}{3}} = 3a^1 = 3a \)
2. \( 2a^{\frac{1}{3}} \cdot a^{\frac{5}{3}} = 2a^{\left(\frac{1}{3} + \frac{5}{3}\right)} = 2a^{\frac{6}{3}} = 2a^2 \)

### Step 5: Final Simplified Expression
Now, combine the simplified terms:

\[
3a + 2a^2
\]

Thus, the final simplified expression is:

\[
2a^2 + 3a
\]

Would you like further details or clarifications on any step?

Here are 5 related questions you might consider:
1. How do exponent rules work for fractional exponents?
2. How can we handle negative exponents in more complex expressions?
3. What happens when you multiply terms with different bases and exponents?
4. Can we factor the final expression \( 2a^2 + 3a \)?
5. How can this technique be applied to more advanced algebraic simplifications?

**Tip**: Always simplify terms involving exponents first, as this can help avoid unnecessary complexity in the solution.

Simplify the expression \( \frac{3a^{-\frac{2}{3}} + 2a^{\frac{1}{3}}}{a^{-\frac{5}{3}}} \).

Learn how to simplify an algebraic expression involving fractional exponents: \( \frac{3a^{-\frac{2}{3}} + 2a^{\frac{1}{3}}}{a^{-\frac{5}{3}}} \). This solution covers the application of exponent rules and algebraic distribution. Suitable for high school students in Grades 10-12.

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