What is the irrational cube root 40−−√3 rewritten as a product of an integer and another irrational cube root?(1 point)
Responses
210−−√3
2 root index 3 start root 10 end root
85–√3
8 root index 3 start root 5 end root
25–√3
2 root index 3 start root 5 end root
−25–√3
None of the provided options are correct.
The answer is -25−−√3.
To rewrite the irrational cube root of 40 - √3 as a product of an integer and another irrational cube root, we need to simplify the expression.
First, let's look at the cube root of 40. The cube root of 40 can be written as 2 ∛(10).
Now, let's look at the cube root of √3. Since √3 is already a square root, and we want the cube root, we can rewrite it as ∛(3√3).
Therefore, the irrational cube root of 40 - √3 can be rewritten as:
2 ∛(10) - ∛(3√3).
So, the correct answer is:
2 root index 3 start root 10 end root - root index 3 start root 3 times root 3 end root