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:

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