ewrite the irrational cube root 48−−√3 as a product of an integer and another irrational cube root.(1 point) Responses 412−−√3 4 root index 3 start root 12 end root −26–√3 negative 2 root index 3 start root 6 end root 43–√3 4 root index 3 start root 3 end root 26–√3
The correct response is:
4 root index 3 start root 3 end root −26–√3
To rewrite the irrational cube root ∛48√3 as a product of an integer and another irrational cube root, we can simplify it as ∛48√3 = ∛(16 × 3)√3 = ∛16√3 × ∛3 = 2√3 × ∛3. Therefore, the answer is 2√3.
To rewrite the irrational cube root ∛48√3 as a product of an integer and another irrational cube root, we need to look for a perfect cube that is a factor of 48.
First, let's find the prime factorization of 48:
48 = 2^4 * 3
Since we are looking for a perfect cube, we can break down 48 into its prime factors in groups of three (2^3 * 2 * 3).
Now, rewrite the cube root expression:
∛48√3 = ∛(2^3 * 2 * 3)√3
Using the properties of radicals, we can split the radical into two separate radicals:
= ∛(2^3)∛(2 * 3)√3
Simplifying each radical:
= 2∛6√3
Therefore, the irrational cube root ∛48√3 can be written as the product 2∛6√3, where 2 is an integer and ∛6√3 is another irrational cube root.