To rewrite the cube root of 48 as the product of an integer and another irrational cube root, we first factor 48 into its prime factors.
The prime factorization of 48 is: \[ 48 = 16 \times 3 = 4^2 \times 3 = 2^4 \times 3. \]
Now, we can express the cube root of 48 as follows: \[ \sqrt[3]{48} = \sqrt[3]{2^4 \times 3}. \]
Using the property of cube roots, we can separate the factors: \[ \sqrt[3]{48} = \sqrt[3]{2^3 \times 2 \times 3} = \sqrt[3]{2^3} \times \sqrt[3]{2 \times 3}. \]
Since \(\sqrt[3]{2^3} = 2\), we have: \[ \sqrt[3]{48} = 2 \times \sqrt[3]{6}. \]
Thus, the cube root of 48 can be rewritten as: \[ \sqrt[3]{48} = 2\sqrt[3]{6}. \]