What is the irrational cube root −24−−−−√3 rewritten as a product of an integer and another irrational cube root?

1 answer

To rewrite the expression \(-\sqrt[3]{24}\) as a product of an integer and another cube root, we can manipulate the expression as follows:

  1. Recognize that \(24\) can be factored into its prime factors. We have: \[ 24 = 8 \times 3 \] where \(8\) is \(2^3\).

  2. Rewriting the cube root of \(24\): \[ -\sqrt[3]{24} = -\sqrt[3]{8 \times 3} \] Using the property of cube roots that states \(\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}\), we can split the cube root: \[ -\sqrt[3]{8 \times 3} = -\sqrt[3]{8} \times \sqrt[3]{3} \]

  3. Knowing that \(\sqrt[3]{8} = 2\) since \(8 = 2^3\): \[ -\sqrt[3]{8} \times \sqrt[3]{3} = -2 \times \sqrt[3]{3} \]

Thus, we can rewrite \(-\sqrt[3]{24}\) as: \[ -\sqrt[3]{24} = -2\sqrt[3]{3} \]

So, the final rewritten form of \(-\sqrt[3]{24}\) is: \[ -2\sqrt[3]{3} \]