Question

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

4^3√3
4^3√3
2^3√6
-2^3√6

Answer 1

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

We have:
\[
48 = 16 \times 3 = 4^2 \times 3
\]

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

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

Since \(16 = 4^2\), we have:

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

Thus,

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

However, this can be adjusted further for clarity. More straightforwardly, we find another approach to split it.

We can see:
\[
48 = 8 \times 6
\]

And since \(\sqrt[3]{8} = 2\):

\[
\sqrt[3]{48} = \sqrt[3]{8 \times 6} = \sqrt[3]{8} \cdot \sqrt[3]{6} = 2 \cdot \sqrt[3]{6}
\]

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

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

This results in:

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

So the final answer is:

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