What is the irrational cube root 3√81−−

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 answer

To rewrite the expression \( \sqrt[3]{81} \) as a product of an integer and another irrational cube root, we need to factor \( 81 \) into a perfect cube and a remaining factor.

First, we recognize that \( 81 = 27 \times 3 \), where \( 27 \) is a perfect cube (\( 27 = 3^3 \)).

Now we can rewrite the cube root:

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

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

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

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

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

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

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