Products of Irrational Square Roots Quick Check

Question

3 of 53 of 5 Items
Question
Rewrite the irrational cube root ^3√48
as a product of an integer and another irrational cube root.(1 point)
Responses

2^3√6
2 root index 3 start root 6 end root

4^3√3
4 root index 3 start root 3 end root

−2^3√6
negative 2 root index 3 start root 6 end root

4^3√12

Answer 1

To rewrite the irrational cube root \( \sqrt[3]{48} \) as a product of an integer and another irrational cube root, we can factor \( 48 \) into its prime factors.

The prime factorization of \( 48 \) is:

\[ 48 = 16 \times 3 = 4^2 \times 3 = 2^4 \times 3 \]

We can express \( 48 \) as \( 8 \times 6 \), where \( 8 \) is a perfect cube. Therefore,

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

So, the correct answer is:

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

This corresponds to the answer choice:

2^3√6 or 2 root index 3 start root 6 end root.