To rewrite the irrational cube root \(-\sqrt[3]{81}\) as a product of an integer and another irrational cube root, we start by simplifying \(\sqrt[3]{81}\).
First, we recognize that \(81\) can be expressed in terms of its prime factors: \[ 81 = 3^4 \]
Now we can find the cube root: \[ \sqrt[3]{81} = \sqrt[3]{3^4} \] Using the property of cube roots, this can be simplified: \[ \sqrt[3]{3^4} = \sqrt[3]{3^3 \cdot 3^1} = \sqrt[3]{3^3} \cdot \sqrt[3]{3^1} \] Calculating the cube root of \(3^3\) gives \(3\): \[ \sqrt[3]{3^3} = 3 \] Thus, we can express \(\sqrt[3]{81}\) as: \[ \sqrt[3]{81} = 3 \cdot \sqrt[3]{3} \]
Now substituting this back into our expression, we get: \[ -\sqrt[3]{81} = - (3 \cdot \sqrt[3]{3}) = -3 \cdot \sqrt[3]{3} \]
Therefore, \(-\sqrt[3]{81}\) rewritten as a product of an integer and another irrational cube root is: \[ \boxed{-3 \sqrt[3]{3}} \]