The problem is to simplify the cube root of $24x^2y^3$, written as:

$\sqrt[3]{24x^2y^3}$

Step 1: Factor inside the cube root

Break down $24x^2y^3$ into prime factors and powers: $24 = 2^3 \times 3$ So: $24x^2y^3 = (2^3) \times 3 \times x^2 \times y^3$

Step 2: Apply the cube root

Now, apply the cube root to each term: [ \sqrt[3]{24x^2y^3} = \sqrt[3]{2^3} \times \sqrt[3]{3} \times \sqrt[3]{x^2} \times \sqrt[3]{y^3} ]

Simplify the terms:

• $\sqrt[3]{2^3} = 2$
• $\sqrt[3]{y^3} = y$
• The other terms ($\sqrt[3]{3}$ and $\sqrt[3]{x^2}$) remain under the cube root.

Final Answer:

$2y\sqrt[3]{3x^2}$

Let me know if you need further clarification or details!

Related Questions:

1. How do you simplify higher-order roots?
2. What are the differences between square and cube roots?
3. Can cube roots simplify further if we add or subtract terms?
4. How does cube root simplification work in fractional exponents?
5. Why do cube roots work differently on odd and even powers?

Tip: Always start by breaking down numbers into their prime factors when simplifying roots!