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

43–√3
4 root index 3 start root 3 end root

26–√3

Answer 1

To rewrite \( \sqrt[3]{48} \) as a product of an integer and another irrational cube root, we can first factor 48 into \( 24 \times 2 \):

\[
\sqrt[3]{48} = \sqrt[3]{24 \times 2} = \sqrt[3]{24} \cdot \sqrt[3]{2}
\]

Next, we can factor 24 further into \( 8 \times 3 \):

\[
\sqrt[3]{24} = \sqrt[3]{8 \times 3} = \sqrt[3]{8} \cdot \sqrt[3]{3}
\]

Since \( \sqrt[3]{8} = 2 \), we have:

\[
\sqrt[3]{24} = 2 \cdot \sqrt[3]{3}
\]

Putting it all together:

\[
\sqrt[3]{48} = 2 \cdot \sqrt[3]{24} = 2 \cdot 2 \cdot \sqrt[3]{3} \cdot \sqrt[3]{2} = 4 \cdot \sqrt[3]{6}
\]

Thus, we can express \( \sqrt[3]{48} \) as:

\[
4 \sqrt[3]{6}
\]

Among the given response options, the correct answer is:

**412−−√3 \ (4 \sqrt[3]{12})**.