what is the irrational cube root ^3 radical 81 rewritten as a product of an integer and another irrational cube root

1 answer

To rewrite the expression \( \sqrt[3]{81} \) as a product of an integer and another cube root, we can first factor \( 81 \).

Since \( 81 = 27 \times 3 \) and \( 27 \) is a perfect cube (\( 27 = 3^3 \)), we can proceed as follows:

\[ \sqrt[3]{81} = \sqrt[3]{27 \times 3} \]

Using the property of cube roots that states \( \sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b} \), we can rewrite this as:

\[ \sqrt[3]{81} = \sqrt[3]{27} \times \sqrt[3]{3} \]

Since \( \sqrt[3]{27} = 3 \), we substitute this back in:

\[ \sqrt[3]{81} = 3 \times \sqrt[3]{3} \]

Therefore, \( \sqrt[3]{81} \) rewritten as a product of an integer and another irrational cube root is:

\[ 3 \sqrt[3]{3} \]