Question

Rewrite the irrational cube root ^3√48 as a product of an integer and another irrational cube root.

a. 2^3√6
b. 4^3√3
c. 4^3√12
d. -2^3√6

Answer 1

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

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

Now we can express \( \sqrt[3]{48} \) as follows:

\[
\sqrt[3]{48} = \sqrt[3]{16 \cdot 3} = \sqrt[3]{16} \cdot \sqrt[3]{3}
\]

Since \( \sqrt[3]{16} = \sqrt[3]{2^4} = \sqrt[3]{2^3 \cdot 2^1} = 2\sqrt[3]{2} \), we can rewrite that as:

\[
\sqrt[3]{48} = 2\sqrt[3]{16} \cdot \sqrt[3]{3} = 2 \cdot \sqrt[3]{16 \cdot 3} = 2\sqrt[3]{48}
\]

So, instead of \( 2 \sqrt[3]{3} \), we can try rewriting it as:

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

The forms you've given are different options, but you'll find that:

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

This corresponds to:

b. \( 4\sqrt[3]{3} \)

So the correct answer is:

**b. \( 4\sqrt[3]{3} \)**.