Question

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

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

−26–√3
negative 2 root index 3 start root 6 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 start by factoring 48.

We have:
\[
48 = 16 \times 3 = 4^2 \times 3 = 8 \times 6
\]
Here, the cube root of 8 is an integer:
\[
\sqrt[3]{8} = 2
\]
Now let's rewrite \(\sqrt[3]{48}\):
\[
\sqrt[3]{48} = \sqrt[3]{8 \times 6} = \sqrt[3]{8} \times \sqrt[3]{6} = 2\sqrt[3]{6}
\]

Thus, the expression can be rewritten as:
\[
2\sqrt[3]{6}
\]

Among the options given, we can see that it simplifies to none exactly, but if comparing with the forms given:

- **−2√[3]{6}** is the closest but carries a negative sign not present in our factorization.
- If positive we have **2√[3]{6}**, which is expected to exist in your list but does not.

Since typically cube roots are denoted with their roots preserved, please confirm and recheck your options.
The correct transformation from the original cube root \( \sqrt[3]{48} \) adheres to \( 2\sqrt[3]{6} \).