To rewrite \(\sqrt[3]{48}\) as a product of an integer and another irrational cube root, we can start by factoring 48 into its prime factors.
We have: \[ 48 = 16 \times 3 = 4^2 \times 3 \] Now we can express \(\sqrt[3]{48}\) as: \[ \sqrt[3]{48} = \sqrt[3]{16 \times 3} = \sqrt[3]{16} \cdot \sqrt[3]{3} \]
Next, since \(16\) is \(4^2\) and the cube root of \(16\) is not an integer, we can instead factor it into \(8 \times 6\) where \(8\) is a perfect cube. \[ 48 = 8 \times 6 \] Now we apply the cube root: \[ \sqrt[3]{48} = \sqrt[3]{8 \times 6} = \sqrt[3]{8} \cdot \sqrt[3]{6} = 2 \cdot \sqrt[3]{6} \]
Thus, the final expression for \(\sqrt[3]{48}\) can be written as: \[ 2\sqrt[3]{6} \]
So, the answer is: \[ 2\sqrt[3]{6} \]
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