The question asks to rewrite the irrational cube root $\sqrt[3]{-24}$ as a product of an integer and another irrational cube root.

Step-by-Step Solution:

1. Expression Analysis: We are working with $\sqrt[3]{-24}$. Since $-24$ is negative, its cube root will also be negative. Let's factorize $-24$: $-24 = -8 \cdot 3$

2. Cube Root Separation: Using the property of cube roots, $\sqrt[3]{a \cdot b} = \sqrt[3]{a} \cdot \sqrt[3]{b}$: [ \sqrt[3]{-24} = \sqrt[3]{-8} \cdot \sqrt[3]{3} ]

3. Simplify the Cube Root of $-8$: The cube root of $-8$ is $-2$ because: $(-2)^3 = -8$

4. Final Answer: Substituting back: [ \sqrt[3]{-24} = -2 \cdot \sqrt[3]{3} ]

Thus, the correct answer is: $\boxed{-2 \sqrt[3]{3}}$

Let me know if you have questions or need further clarification.

Related Questions:

1. What is the cube root of a negative number generally?
2. How do you simplify cube roots of larger numbers?
3. What is the difference between square roots and cube roots?
4. How would you simplify $\sqrt[3]{54}$ into a similar form?
5. Can all irrational cube roots be expressed as a product of an integer and a cube root?

Tip:

Remember that the cube root of a negative number will always result in a negative value, unlike square roots, which are undefined for negatives in the real numbers.