Evaluate.

\[\dfrac{3^{^{\scriptsize\dfrac{1}5}}}{\sqrt[5]{-96}}=\]

The expression to evaluate is:

\[
\frac{3^{\frac{1}{5}}}{\sqrt[5]{-96}}.
\]

### Step 1: Simplify the denominator
The denominator involves the fifth root of \(-96\), which can be written as:

\[
\sqrt[5]{-96} = (-96)^{\frac{1}{5}}.
\]

The fifth root of a negative number is a real number, since odd roots of negative numbers are real. Thus:

\[
(-96)^{\frac{1}{5}} = -96^{\frac{1}{5}}.
\]

### Step 2: Rewrite the expression
Now, the original expression becomes:

\[
\frac{3^{\frac{1}{5}}}{-96^{\frac{1}{5}}} = 3^{\frac{1}{5}} \times (-96^{-\frac{1}{5}}).
\]

### Step 3: Combine the powers
We now combine the two terms, both raised to powers of \( \frac{1}{5} \). This results in:

\[
= \left( \frac{3}{96} \right)^{\frac{1}{5}} = \left( \frac{1}{32} \right)^{\frac{1}{5}}.
\]

Since \( \frac{1}{32} = 2^{-5} \), we get:

\[
= (2^{-5})^{\frac{1}{5}} = 2^{-1}.
\]

### Step 4: Final simplification
The expression simplifies to:

\[
2^{-1} = \frac{1}{2}.
\]

Thus, the value of the expression is:

\[
\boxed{\frac{1}{2}}.
\]

Would you like more details on any of the steps, or do you have any questions?

### Here are some related questions to deepen your understanding:
1. How would you compute \( 3^{1/5} \) numerically?
2. What is the general rule for finding the fifth root of a negative number?
3. Can you simplify \( \frac{2^{1/3}}{4^{1/6}} \)?
4. How would the expression change if the exponent in the denominator were \( \frac{1}{3} \)?
5. Can the same approach be used for square roots of negative numbers?

#### Tip: When working with fractional exponents, always consider if you can rewrite the base in terms of smaller powers, especially if simplifying roots.

\[\dfrac{3^{\scriptsize \dfrac{1}{5}}}{\sqrt[5]{-96}} = \]

Learn how to evaluate the expression 3^(1/5) divided by the fifth root of -96 with a detailed step-by-step solution. Explore key mathematical concepts like exponents and roots of negative numbers, along with their properties. Suitable for high school students in Grades 9-12.

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