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 negative 2 root index 3 start root 5 end root 85–√3 8 root index 3 start root 5 end root 25–√3 2 root index 3 start root 5 end root 210−−√3
To rewrite the irrational cube root √[3](40) as a product of an integer and another irrational cube root, we need to find a perfect cube that divides 40.
The largest perfect cube that divides 40 is 8, which is equal to 2^3.
Thus, we can rewrite the irrational cube root √[3](40) as 2√[3](5).
Therefore, the correct response is: 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 the expression as follows:
∛40-√3 = ∛(40 × 1)-√(3 × 1)
Now we can separate the terms inside the cube root and square root:
= ∛(8 × 5) - √(√(3^2))
= ∛8 × ∛5 - √3
The simplified expression 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 need to simplify it.
Step 1: Simplify the cube root of 40√3.
To find a simplified form, we can factorize 40√3.
40√3 = 2 * 2 * 10√3 = 2 * 2 * √(3^3 * 10) = 2 * 2 * √(27 * 10) = 2 * 2 * √270
Step 2: Simplify the square root of 270.
To simplify the square root of 270, we can look for perfect square factors.
270 = 27 * 10 = 3^3 * 2 * 5
Therefore, √270 = √(3^3 * 2 * 5) = √(3^3) * √(2 * 5) = 3√2√5 = 3√10
Step 3: Combine the simplified expressions from steps 1 and 2.
We now have the simplified form of the irrational cube root of 40√3:
40√3 = 2 * 2 * √270 = 2 * 2 * 3√10
Therefore, the irrational cube root of 40√3 can be rewritten as a product of an integer and another irrational cube root as:
2 * 2 * 3√10 = 12√10