Question Rewrite the irrational cube root 48−−√3 as a product of an integer and another irrational cube root.(1 point) Responses 26–√3 2 root index 3 start root 6 end root 43–√3 4 root index 3 start root 3 end root −26–√3 negative 2 root index 3 start root 6 end root 412−−√3
To rewrite the irrational cube root of 48√3 as a product of an integer and another irrational cube root, we can simplify it as follows:
48√3 = 2 √(16 × 3) = 2 √(4 × 4 × 3) = 2 × 4√3 = 8√3
Therefore, the irrational cube root of 48√3 can be written as the product of 8 and the cube root of 3.
To rewrite the irrational cube root of 48√3 as a product of an integer and another irrational cube root, we need to simplify it.
Step 1: Split 48√3 into two parts
Let's first split 48√3 into the product of two numbers: one that is a perfect cube and one that is not a perfect cube. In this case, 48 can be written as 2^4 * 3, and since 3 is not a perfect cube, we can rewrite it as (2^3 * 3) * 2√3.
Step 2: Rewrite the expression
Now, we can rewrite the expression as a product of an integer and another irrational cube root. Apply the property of cube roots, which states that ∛(a * b) = ∛a * ∛b.
The expression becomes:
(2∛2 * ∛3) * ∛2√3
Simplified, this is:
2∛2 * 3∛2 * ∛(√3)
Therefore, the irrational cube root of 48√3 can be rewritten as the product:
6∛2 * ∛(√3)