Rewrite the irrational cube root 48−−√3

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 \(2 \sqrt[3]{6}\)).