What is the irrational cube root ^3^√81 rewritten as a product of an integer and another irrational cube root?

1 answer

To rewrite \( \sqrt[3]{81} \) as a product of an integer and another irrational cube root, we start by factoring \( 81 \).

We can express \( 81 \) as \( 27 \times 3 \), since \( 27 = 3^3 \) is a perfect cube.

Now, we can rewrite the cube root:

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

Using the property of cube roots, we can separate the product:

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

Since \( \sqrt[3]{27} = 3 \):

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

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

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