To rewrite \( \sqrt[3]{48} \) as a product of an integer and another irrational cube root, we can factor 48 into a product of a perfect cube and another integer.
The prime factorization of 48 is:
\[ 48 = 16 \times 3 = 2^4 \times 3 \]
We can also express 16 as \( 2^4 \), which is \( (2^3 \times 2) = 8 \times 6 \).
Thus:
\[ \sqrt[3]{48} = \sqrt[3]{8 \times 6} = \sqrt[3]{8} \cdot \sqrt[3]{6} \]
Since \( \sqrt[3]{8} = 2 \), we have:
\[ \sqrt[3]{48} = 2 \cdot \sqrt[3]{6} \]
Hence, the final answer is:
2 \( \sqrt[3]{6} \)
None of the provided options seem to match this answer directly. If you still want to complete the exercise using the choices given, you may need to clarify or check the problem context again.