What is the irrational cube root 81−−√3 rewritten as a product of an integer and another irrational cube root? Note: You will find the option to add a cubic root symbol in the Algebra ( × ) keyboard. (1 point)

Answer 1

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

First, notice that:

\[ 81 = 3^4 \]

Now we can use the properties of cube roots:

\[ \sqrt[3]{81} = \sqrt[3]{3^4} \]

This can be broken down using the property \( \sqrt[3]{a^m} = a^{m/3} \):

\[ \sqrt[3]{3^4} = 3^{4/3} = 3^{1 + 1/3} = 3 \cdot 3^{1/3} \]

Thus:

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

So, we can rewrite \( \sqrt[3]{81} \) as:

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

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

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