What is the irrational cube root 40−−√3 rewritten as a product of an integer and another irrational cube root?(1 point)
Responses
2^3√5
2 root index 3 start root 5 end root
8^3√5
8 root index 3 start root 5 end root
−2^3√5
negative 2 root index 3 start root 5 end root
2^3√10
None of the given options are correct. The correct answer is: 2∛(10)
To rewrite the irrational cube root ∛(40-√3) as a product of an integer and another irrational cube root, we need to find a perfect cube that can be factored out of 40-√3.
Since 40 is a perfect cube of 2^3 * 5, we can rewrite it as (2^3) * (5-√3).
Therefore, the irrational cube root ∛(40-√3) can be rewritten as 2∛(5-√3).
To rewrite the given irrational cube root as a product of an integer and another irrational cube root, we need to find a perfect cube that divides 40 and then express the remaining factor as another irrational cube root.
Let's find the largest perfect cube that divides 40. The largest perfect cube less than or equal to 40 is 8 (2^3).
Now we can rewrite the irrational cube root of 40 as a product:
(2^3√5)
So the correct answer is 2 root index 3 start root 5 end root.
Note: The other options provided, 8^3√5 and −2^3√5, are not correct because the cube root of 40 cannot be written as a perfect cube multiplied by an irrational cube root.