Rewrite the irrational cube root ³√48 as a product of an interrogator and another irrational square root
³√48 can be rewritten as ³√(16 * 3) = ³√16 * ³√3 = 2 * ³√3
Therefore, the irrational cube root ³√48 can be expressed as the product of the interrogator 2 and the irrational square root ³√3.
To rewrite the irrational cube root ³√48 as a product of an interrogator and another irrational square root, we can start by simplifying 48 into its prime factors.
The prime factorization of 48 is: 2^4 * 3.
Now let's consider the cube root of 48, which is represented as ³√48.
Using the prime factorization, we can rewrite 48 as (2^4 * 3). Taking the cube root of each of the factors separately, we get:
³√(2^4 * 3)
Using the property of exponents, we can rewrite this as:
³√2^4 * ³√3
Since the cube root of 2^4 is 2, we can simplify this further:
2 * ³√3
Now we have expressed the irrational cube root ³√48 as a product of an interrogator (2) and another irrational square root (√3).
So, ³√48 can be written as 2 * √3.
To rewrite the irrational cube root ³√48 as a product of an interrogator and another irrational square root, we need to simplify it first.
Step 1: Prime factorize 48: 48 = 2 * 2 * 2 * 2 * 3 = 2^4 * 3
Step 2: Identify the perfect cubes that divide 48:
- The largest perfect cube that divides 48 is 2^3 = 8
Step 3: Rewrite 48 using the largest perfect cube:
48 = 8 * (2^4 * 3 / 8) = 8 * (2 * 2^3 * 3 / 8) = 8 * (2 * ³√2^3 * ³√3)
Therefore, the irrational cube root ³√48 can be rewritten as:
³√48 = 8 * ³√2 * ³√3
So, the product of an interrogator and another irrational square root is 8 * ³√2.