What is the irrational cube root 40−−√3 rewritten as a product of an integer and another irrational cube root?(1 point) Responses 85–√3 8 root index 3 start root 5 end root −25–√3 negative 2 root index 3 start root 5 end root 210−−√3 2 root index 3 start root 10 end root 25–√3
The irrational cube root of 40 can be rewritten as 2 root(3)(10).
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 = ∛(8*5)-√3 = ∛8*∛5-√3 = 2∛5-√3
Therefore, the answer is 2∛5-√3.
To rewrite the irrational cube root of 40-√3 as a product of an integer and another irrational cube root, we can follow these steps:
Step 1: Simplify the perfect cube roots:
The cube root of 40 is the cube root of 8 multiplied by the cube root of 5 (since 8^3 = 512 and 5^3 = 125).
The cube root of √3 can be simplified as the cube root of 3 multiplied by the cube root of √1 (since √3 = √(3 * 1) = √3 * √1).
So the given expression can be simplified as:
cube root of (8 * 5) - cube root of (3 * √1)
Step 2: Rewrite the expression:
The cube root of 8 can be written as 2, since 2^3 = 8.
The cube root of 5 can be left as it is, since it is an irrational number.
The cube root of √1 can be written as √(√1) = √1, which is equivalent to 1.
So the expression becomes:
(2 * cube root of 5) - (1 * cube root of 3)
Therefore, the irrational cube root of 40-√3 can be rewritten as:
2 * cube root of 5 - cube root of 3