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

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

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

26–√3

Answer 1

To rewrite the irrational cube root \(\sqrt[3]{48}\) as a product of an integer and another irrational cube root, we start by factoring \(48\).

First, notice that:
\[
48 = 16 \times 3 = 2^4 \times 3
\]

Now, we can find the cube root:
\[
\sqrt[3]{48} = \sqrt[3]{16 \times 3} = \sqrt[3]{16} \times \sqrt[3]{3}
\]

Since \(16 = 2^4\), we can take out a perfect cube:
\[
\sqrt[3]{16} = \sqrt[3]{2^3 \times 2} = 2 \sqrt[3]{2}
\]

Putting it all together, we have:
\[
\sqrt[3]{48} = 2 \sqrt[3]{2} \times \sqrt[3]{3} = 2 \sqrt[3]{6}
\]

So the original expression can be rewritten as:
\[
\sqrt[3]{48} = 2 \sqrt[3]{6}
\]

Based on the given options, the correct choice that matches our result is:
\[
-2 \sqrt[3]{6}
\]

However, there may have been a typo in the listed choices. Assuming the correct reinterpretation, the answer would be:
\[
2 \sqrt[3]{6}
\]

Ensure to check the answer options provided for exact matches.