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