Question

rewrite the irrational cube root ^3√189 as the product of an integer and another irrational cube root. show your work. Please make it short and simple

Answer 1

To rewrite \(\sqrt[3]{189}\) as the product of an integer and another irrational cube root, we can start by factoring \(189\).

First, we find the prime factorization of \(189\):

\[
189 = 3^3 \times 7
\]

Now, we can express \(\sqrt[3]{189}\) using this factorization:

\[
\sqrt[3]{189} = \sqrt[3]{3^3 \times 7}
\]

We can separate the cube root:

\[
\sqrt[3]{189} = \sqrt[3]{3^3} \times \sqrt[3]{7}
\]

Since \(\sqrt[3]{3^3} = 3\), we have:

\[
\sqrt[3]{189} = 3 \times \sqrt[3]{7}
\]

Thus, \(\sqrt[3]{189}\) can be rewritten as:

\[
\sqrt[3]{189} = 3\sqrt[3]{7}
\]

This is the final result.

Answer 2

Again

Answer 3

Sure! Let's rewrite \(\sqrt[3]{189}\) as the product of an integer and another irrational cube root.

1. **Factor \(189\)**:
\[
189 = 3^3 \times 7
\]

2. **Write the cube root**:
\[
\sqrt[3]{189} = \sqrt[3]{3^3 \times 7}
\]

3. **Separate the cube roots**:
\[
\sqrt[3]{189} = \sqrt[3]{3^3} \times \sqrt[3]{7}
\]

4. **Simplify**:
\[
\sqrt[3]{3^3} = 3
\]
Thus,
\[
\sqrt[3]{189} = 3 \times \sqrt[3]{7}
\]

So, the final expression is:
\[
\sqrt[3]{189} = 3\sqrt[3]{7}
\]