Question

Rewrite the irrational cube root 48−−√3 as a product of an integer and another irrational cube root.(1 point)
Responses

−26–√3
negative 2 root index 3 start root 6 end root

412−−√3
4 root index 3 start root 12 end root

26–√3
2 root index 3 start root 6 end root

43–√3

Answer 1

To rewrite the irrational cube root \( \sqrt[3]{48} \) as a product of an integer and another cube root, we can first factor \( 48 \).

Since \( 48 = 16 \times 3 = 4^2 \times 3 \), we can break it down like this:

\[
\sqrt[3]{48} = \sqrt[3]{16 \times 3} = \sqrt[3]{16} \times \sqrt[3]{3}
\]

Now, we also know that \( 16 = 2^4 \), which can be expressed with a cube root:

\[
\sqrt[3]{16} = \sqrt[3]{2^4} = 2 \cdot \sqrt[3]{2}
\]

Thus we can simplify this further:

\[
\sqrt[3]{48} = 2 \cdot \sqrt[3]{16} \cdot \sqrt[3]{3} = 2 \cdot \sqrt[3]{48}
\]

We can also reformulate it as follows:

\[
\sqrt[3]{48} = 4 \cdot \sqrt[3]{3} = 2 \cdot \sqrt[3]{12}
\]

So, in the context of the question, you can write:

\[
\sqrt[3]{48} = 4 \sqrt[3]{12}
\]

Thus, the correct answer from the options you provided is:

**4 \(\sqrt[3]{12}\)** which corresponds to the response **\( 4 \sqrt[3]{12} \)**.