What is the irrational cube root ^3√48 rewritten as a product of an integer and another irrational cube root?
To find an integer and another irrational cube root equivalent to ^3√48, we can express 48 as a product of its prime factors:
48 = 2 * 2 * 2 * 2 * 3
Now, we look for cubes among these prime factors. In this case, we have two cubes: 2 * 2 * 2 = (2^3) = 8.
So, we can rewrite ^3√48 as ^3√(8 * 2 * 3).
Taking the cube root of 8 gives us 2:
^3√(8 * 2 * 3) = 2 * ^3√(2 * 3)
Therefore, the irrational cube root ^3√48 can be rewritten as the product 2 * ^3√(6).
To rewrite the cube root ^3√48 as a product of an integer and another irrational cube root, we need to find a perfect cube that divides 48.
Since 48 is not a perfect cube, we can factorize it to find a perfect cube within it.
The prime factorization of 48 is: 2 * 2 * 2 * 2 * 3.
Now, we can group the factors into perfect cubes:
48 = (2 * 2 * 2) * (2 * 3)
Simplifying further:
48 = (2^3) * (6)
Now we can rewrite the cube root ^3√48 as:
^3√48 = ^3√(2^3 * 6)
= 2 * ^3√6
Therefore, the irrational cube root ^3√48 can be rewritten as the product of the integer 2 and the irrational cube root ^3√6.
To rewrite ^3√48 as a product of an integer and another irrational cube root, we need to find its prime factorization.
First, let's find the prime factors of 48. 48 can be expressed as 2 * 2 * 2 * 2 * 3.
Now, let's group the prime factors of 48 into groups of three. We have 2 * 2 * 2 and 2 * 3.
Since 2 * 2 * 2 equals 8, we can rewrite ^3√48 as ^3√(8 * 2 * 3).
Taking the cube root of 8, we get ^3√8 = 2.
Thus, we can write ^3√48 as 2 * ^3√(2 * 3).
Therefore, the irrational cube root ^3√48 can be rewritten as the product of an integer 2 and another irrational cube root ^3√(2 * 3).