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

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

412−−√3

Answer 1

To rewrite the expression \( \sqrt[3]{48} \) as a product of an integer and another irrational cube root, we first factor \( 48 \) into its prime factors:

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

Now we can rewrite the cube root:

\[
\sqrt[3]{48} = \sqrt[3]{16 \times 3} = \sqrt[3]{16} \times \sqrt[3]{3} = \sqrt[3]{4^2} \times \sqrt[3]{3} = \sqrt[3]{4^2} \times \sqrt[3]{3}
\]

We know that \( \sqrt[3]{4^2} = \sqrt[3]{16} \), which can be rewritten as \( 4^{2/3} = 4^{1} \cdot 4^{-1/3} = 4 \cdot \sqrt[3]{4} \).

Now rewriting it gives us:

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

So the correct response is:

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

Thus, the correct option matching this is:

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