To rewrite \( \sqrt[3]{48} \) as a product of an integer and another irrational cube root, we need to factor \( 48 \) into its prime factors.
We can express \( 48 \) as follows:
\[ 48 = 16 \times 3 = 4^2 \times 3 = 2^4 \times 3 \]
Now, we can take the cube root:
\[ \sqrt[3]{48} = \sqrt[3]{16 \times 3} = \sqrt[3]{16} \cdot \sqrt[3]{3} \]
Next, simplify \( \sqrt[3]{16} \):
\[ \sqrt[3]{16} = \sqrt[3]{2^4} = 2^{4/3} = 2^{1 + 1/3} = 2 \cdot \sqrt[3]{2} \]
Substituting this back into the expression, we get:
\[ \sqrt[3]{48} = 2 \cdot \sqrt[3]{2} \cdot \sqrt[3]{3} = 2 \cdot \sqrt[3]{6} \]
Therefore, we can write:
\[ \sqrt[3]{48} = 4 \cdot \sqrt[3]{6} \]
Thus, the rational cube root of \( 48 \) is rewritten as:
\[ \sqrt[3]{48} = 4 \cdot \sqrt[3]{6} \]