The expression in the image is:

$\left( \frac{-64c^6}{a^9b^{\frac{1}{2}}} \right)^{\frac{1}{3}}$

Let's simplify this step-by-step.

Step 1: Apply the exponent $\frac{1}{3}$ to both the numerator and the denominator:

$\left( \frac{-64c^6}{a^9b^{\frac{1}{2}}} \right)^{\frac{1}{3}} = \frac{\left( -64c^6 \right)^{\frac{1}{3}}}{\left( a^9b^{\frac{1}{2}} \right)^{\frac{1}{3}}}$

Step 2: Simplify each part of the fraction:

• Numerator: $(-64c^6)^{\frac{1}{3}} = (-64)^{\frac{1}{3}} \cdot (c^6)^{\frac{1}{3}}$ (-64)^{\frac{1}{3}} = -4 \quad \text{(since the cube root of -64$is$-4)} $(c^6)^{\frac{1}{3}} = c^2$ Thus, the numerator becomes: $-4c^2$

• Denominator: $(a^9b^{\frac{1}{2}})^{\frac{1}{3}} = a^{9 \cdot \frac{1}{3}} \cdot b^{\frac{1}{2} \cdot \frac{1}{3}} = a^3b^{\frac{1}{6}}$

Step 3: Combine the results:

$\frac{-4c^2}{a^3b^{\frac{1}{6}}}$

Thus, the simplified expression is:

$\frac{-4c^2}{a^3b^{\frac{1}{6}}}$

Would you like more details on any step, or do you have any questions?

Here are 5 related questions you could explore:

1. What are the general rules for simplifying expressions with fractional exponents?
2. How do cube roots work for both positive and negative numbers?
3. Can you generalize how to apply exponents to both the numerator and denominator?
4. How would the result change if the exponent was different, say $\frac{1}{2}$?
5. How do you deal with variables raised to fractional powers in real-life applications?

Tip: Always simplify each part of a complex fraction separately before combining them for clarity.