To rewrite the irrational cube root \( \sqrt[3]{81} \) as a product of an integer and another irrational cube root, we start by breaking down \( 81 \) into its prime factors:
\[ 81 = 3^4 \]
Now, we can express the cube root of \( 81 \):
\[ \sqrt[3]{81} = \sqrt[3]{3^4} \]
Using the property of exponents \( \sqrt[3]{a^b} = a^{b/3} \), we can further simplify:
\[ \sqrt[3]{3^4} = 3^{4/3} = 3^{1 + 1/3} = 3^1 \cdot 3^{1/3} = 3 \cdot \sqrt[3]{3} \]
Thus, we can rewrite \( \sqrt[3]{81} \) as:
\[ \sqrt[3]{81} = 3 \cdot \sqrt[3]{3} \]
This expression is now in the desired form, showing \( \sqrt[3]{81} \) as a product of an integer (which is \( 3 \)) and another irrational cube root (\( \sqrt[3]{3} \)).
So, the final answer is:
\[ \sqrt[3]{81} = 3 \cdot \sqrt[3]{3} \]