Question

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

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

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

43–√3
4
3
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 start by factoring 48.

We can express 48 as:

\[
48 = 16 \times 3
\]

The cube root of 16 can be simplified:

\[
\sqrt[3]{16} = \sqrt[3]{4^2} = \sqrt[3]{(2^2)^2} = \sqrt[3]{(2^4)} = 2^{4/3} \text{ (not an integer or a simple cube root)}
\]

It's better to express it as follows:

\[
48 = 8 \times 6
\]

Where \( 8 \) is a perfect cube. Thus:

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

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

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

So the correct expression is:

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

Therefore, the response is:

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