What is the irrational cube root 3^√81 rewritten as a product of an integer and another irrational cube root?
wrong again
∛81 = ∛(27*3) = 3∛3
Apologies for the incorrect response earlier. You are correct.
The irrational cube root of 81, \(\sqrt[3]{81}\), can be rewritten as the product of an integer and another irrational cube root.
\(\sqrt[3]{81} = \sqrt[3]{27 \cdot 3} = 3 \sqrt[3]{3}\)
Thus, the irrational cube root \(3^{\sqrt[3]{81}}\) can be rewritten as the product of an integer (3) and another irrational cube root: \(3 \sqrt[3]{3}\).
To rewrite the irrational cube root \(3^{\sqrt{81}}\) as a product of an integer and another irrational cube root, we need to express \(\sqrt{81}\) in terms of a cube root.
We know that \(\sqrt{81} = 9\) since the square root of a positive number is always positive.
Now, let's rewrite \(3^{\sqrt{81}}\) as \((3^9)^{\frac{1}{9}}\).
Since \(3^9\) is an integer (specifically, \(19683\)), we can rewrite our expression as \(19683^{\frac{1}{9}}\).
Thus, the irrational cube root \(3^{\sqrt{81}}\) can be rewritten as the product of an integer (19683) and another irrational cube root: \(19683^{\frac{1}{9}}\).