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
412−−√3
4 root index 3 start root 12 end root
26–√3
2 root index 3 start root 6 end root
4∛3
To rewrite the irrational cube root ∛48 as a product of an integer and another irrational cube root, we need to factorize the number 48 and express it in terms of its prime factors.
First, we find the prime factorization of 48:
48 = 2 * 2 * 2 * 2 * 3
Now, we group the prime factors in sets of three to form perfect cubes:
48 = (2 * 2 * 2) * (2 * 3)
The first set (2 * 2 * 2) can be simplified as 2^3 = 8.
So, we have:
48 = 8 * (2 * 3)
Simplifying further, we get:
48 = 8 * 6
Therefore, we can rewrite the irrational cube root ∛48 as:
∛48 = ∛(8 * 6)
Now, we can separate the cube root into two parts:
∛48 = ∛8 * ∛6
Simplifying the two cube roots:
∛8 = 2
∛6 remains as it is
Finally, we can rewrite the irrational cube root ∛48 as a product of an integer and another irrational cube root:
∛48 = 2 * ∛6
To rewrite the irrational cube root ∛48 as a product of an integer and another irrational cube root, let's try simplifying it:
∛48
First, let's find the prime factorization of 48:
48 = 2 × 2 × 2 × 2 × 3
Next, we can group the factors in sets of three (since we're looking for a cube root):
∛(2 × 2 × 2 × 2 × 3)
Taking out the cubes from each set gives us:
2 × 2 × ∛(2 × 3)
Simplifying further, we have:
4 ∛(6)
Therefore, we can rewrite the irrational cube root ∛48 as the product of an integer and another irrational cube root as:
4∛6
e
I'm sorry, but "e" is not a valid response in this context. Could you please clarify your response or provide further information?
out of the responses
Out of the responses you provided, the correct answer is:
4∛6