Rewrite the irrational cube root 48−−√3 as a product of an integer and another irrational cube root.(1 point)
The irrational cube root of 48 can be written as a product of an integer and another irrational cube root as 2∛2.
wrong
Apologies for the incorrect answer. Let's find the correct answer.
To express the irrational cube root of 48 as a product of an integer and another irrational cube root, we need to find a cube root that can be simplified from 48.
Since 48 is divisible by 8, we can simplify the cube root of 48 as follows:
√3∛48 = √3∛8 * ∛6 = 2∛2 * ∛6
So, the irrational cube root of 48 can be written as a product of an integer (2) and another irrational cube root (∛2 * ∛6).
To rewrite the irrational cube root ∛48 as a product of an integer and another irrational cube root, you can start by finding the prime factorization of 48.
The prime factorization of 48 is 2^4 * 3^1.
Now, we can rewrite ∛48 as ∛(2^4 * 3^1).
Breaking up the cube root into individual factors, we get: ∛2^4 * ∛3^1.
Simplifying, we have 2 * ∛3 as the final expression.
So, the irrational cube root ∛48 can be written as 2∛3, where 2 is the integer and ∛3 is the other irrational cube root.