To rewrite the cube root of 72, \(\sqrt[3]{72}\), as a product of an integer and another irrational cube root, we can start by factoring 72 into its prime factors.
The prime factorization of 72 is: \[ 72 = 2^3 \times 3^2 \]
Now, we can rewrite \( \sqrt[3]{72} \) using these factors: \[ \sqrt[3]{72} = \sqrt[3]{2^3 \times 3^2} \]
Using the properties of cube roots, we can separate the cube root of the product: \[ \sqrt[3]{72} = \sqrt[3]{2^3} \times \sqrt[3]{3^2} \]
Now, we know that \(\sqrt[3]{2^3} = 2\), so we can simplify: \[ \sqrt[3]{72} = 2 \times \sqrt[3]{3^2} \]
Thus, the cube root of 72 can be expressed as: \[ \sqrt[3]{72} = 2\sqrt[3]{9} \]
So the final answer is: \[ \sqrt[3]{72} = 2\sqrt[3]{9} \]
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