To simplify the given expression:

[ \frac{\sqrt[3]{3x^6}}{\sqrt[3]{5x^4y^3}} ]

we proceed as follows:

Step 1: Break down the cube roots

The expression can be rewritten using the cube root properties:

$\sqrt[3]{\frac{3x^6}{5x^4y^3}}$

Step 2: Simplify the fraction inside the cube root

Combine terms in the numerator and denominator:

$\frac{3x^6}{5x^4y^3} = \frac{3x^{6-4}}{5y^3} = \frac{3x^2}{5y^3}.$

Step 3: Apply the cube root to each part

Use the property of cube roots for fractions:

[ \sqrt[3]{\frac{3x^2}{5y^3}} = \frac{\sqrt[3]{3x^2}}{\sqrt[3]{5y^3}}. ]

Step 4: Simplify each cube root

• $\sqrt[3]{3x^2}$ cannot be simplified further, so it remains as is.
• $\sqrt[3]{y^3} = y$, since the cube root and cube cancel out.

Thus, the expression becomes:

[ \frac{\sqrt[3]{3x^2}}{y\sqrt[3]{5}}. ]

Step 5: Rationalize the denominator

To rationalize the denominator, multiply numerator and denominator by $\sqrt[3]{5^2} = \sqrt[3]{25}$:

[ \frac{\sqrt[3]{3x^2} \cdot \sqrt[3]{25}}{y \cdot \sqrt[3]{5} \cdot \sqrt[3]{25}} = \frac{\sqrt[3]{75x^2}}{y \cdot 5}. ]

Final Simplified Expression:

$\frac{\sqrt[3]{75x^2}}{5y}.$

Let me know if you would like further clarification or details! Here are 5 related questions to explore further:

1. How do cube roots differ from square roots in simplification?
2. What are common methods to rationalize cube root denominators?
3. Can you generalize the process for higher roots like fourth or fifth roots?
4. How would this problem change if exponents were negative?
5. What are practical applications of rationalizing denominators?

Tip: When simplifying expressions with roots, always check for perfect powers in both the numerator and denominator to simplify more efficiently!