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

43–√3
4 root index 3 start root 3 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 start by factoring \( 48 \):

\[
48 = 16 \times 3 = 4^2 \times 3
\]
This can also be expressed as:

\[
48 = 8 \times 6
\]

Thus, we can write:

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

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

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

Therefore, the expression \( \sqrt[3]{48} \) can be rewritten as:

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

The correct response from the options you provided is:

**−26–√3** (interpreted as \(2 \sqrt[3]{6}\)).