Question

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

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

−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

Answer 1

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

The prime factorization of 48 is:

\[
48 = 16 \times 3 = 2^4 \times 3
\]

We can also express 16 as \( 2^4 \), which is \( (2^3 \times 2) = 8 \times 6 \).

Thus:

\[
\sqrt[3]{48} = \sqrt[3]{8 \times 6} = \sqrt[3]{8} \cdot \sqrt[3]{6}
\]

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

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

Hence, the final answer is:

**2 \( \sqrt[3]{6} \)**

None of the provided options seem to match this answer directly. If you still want to complete the exercise using the choices given, you may need to clarify or check the problem context again.