Question

Rewrite the irrational cube root 189−−−√3 as the product of an integer and another irrational cube root. Show your work. Note: You will find the option to add a cubic root symbol in the Algebra ( × ) keyboard

Answer 1

To rewrite the cube root of \(189\) as the product of an integer and another cube root, we need to factor \(189\) in a way that includes a perfect cube.

First, let's factor \(189\):

\[
189 = 3 \times 63
\]
\[
63 = 3 \times 21
\]
\[
21 = 3 \times 7
\]

So, we can write:

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

Now we can use this factorization to rewrite \(\sqrt[3]{189}\):

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

Using the property of cube roots that states \(\sqrt[3]{a \cdot b} = \sqrt[3]{a} \cdot \sqrt[3]{b}\), we can separate the perfect cube from the rest:

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

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

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

Thus, the final product is:

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

This shows that \( \sqrt[3]{189} \) can be expressed as the product of the integer \(3\) and the irrational cube root \( \sqrt[3]{7} \).