Rewrite the irrational cube root 48−−√3 as a product of an integer and another irrational cube root.(1 point) Responses 43–√3 4 root index 3 start root 3 end root −26–√3 negative 2 root index 3 start root 6 end root 26–√3 2 root index 3 start root 6 end root 412−−√3 4 root index 3 start root 12 end root
48−−√3 can be rewritten as 4∛3.
To rewrite the irrational cube root of 48-sqrt(3) as a product of an integer and another irrational cube root, we can factor out a perfect cube from 48.
48 can be expressed as 8 multiplied by 6, and 8 is a perfect cube (2^3 = 8).
So, we can rewrite the expression as follows:
sqrt(48-sqrt(3)) = sqrt(8*6-sqrt(3)) = sqrt(8)*sqrt(6)-sqrt(3) = 2*sqrt(6)-sqrt(3)
Therefore, the irrational cube root of 48-sqrt(3) can be written as 2*sqrt(6)-sqrt(3).
To rewrite the irrational cube root ∛48 as a product of an integer and another irrational cube root, we can look for perfect cube factors of 48.
Step 1: Prime factorize 48: 48 = 2^4 * 3
Step 2: Rewrite the prime factorization using perfect cube factors:
48 = (2^3 * 6) * 3
Step 3: Simplify the expression:
∛48 = ∛(2^3 * 6) * ∛3
= 2∛6 * ∛3
Therefore, the irrational cube root ∛48 can be rewritten as the product 2∛6, where 2 is an integer and ∛6 is another irrational cube root.