What is the irrational cube root 40−−√3 rewritten as a product of an integer and another irrational cube root?(1 point)
Responses
25–√3
2 root index 3 start root 5 end root
210−−√3
2 root index 3 start root 10 end root
85–√3
8 root index 3 start root 5 end root
−25–√3
negative 2 root index 3 start root 5 end root
The correct response is "negative 2 root index 3 start root 5 end root"
To rewrite the irrational cube root ∛40−√3 as a product of an integer and another irrational cube root, we can simplify it as follows:
∛40−√3 = 2∛10−√3
So, the correct answer would be "2√3 start root 10 end root".
To rewrite the irrational cube root ∛(40 - √3) as a product of an integer and another irrational cube root, we can follow these steps:
1. Start by simplifying the expression within the cube root, if possible. In this case, 40 - √3 cannot be simplified further.
2. Express the number inside the cube root as a perfect cube multiplied by another number. We want to find a perfect cube that divides 40. The largest perfect cube that divides 40 is 8 (2^3).
3. Rewrite the expression as a product of an integer and another irrational cube root, using the perfect cube we found in the previous step. In this case, we have:
∛(40 - √3) = ∛(8 * 5 - √3) = ∛(8) * ∛(5 - 1/√3)
Therefore, the irrational cube root ∛(40 - √3) can be rewritten as 2∛(5 - 1/√3), where 2 is the integer and ∛(5 - 1/√3) is the other irrational cube root.