What is the irrational cube root ^3√40 rewritten as a product of an integer and another irrational cube root?
1. 2 ^3√5
2. 2 ^3√10
3. 8 ^3√5
4. -2 ^3√5
To rewrite ^3√40 as a product of an integer and another irrational cube root, we need to find a perfect cube that divides 40.
Let's prime factorize 40: 40 = 2^3 * 5
Since 5 is the only cube factor, we can rewrite ^3√40 as ^3√(8 * 5).
Now, we can rewrite the expression as: 2 ^3√5
Therefore, the correct answer is option 1: 2 ^3√5.
To find the answer, we need to rewrite ^3√40 as a product of an integer and another irrational cube root.
First, let's find the prime factorization of 40:
40 = 2 × 2 × 2 × 5
So, ^3√40 = ^3√(2 × 2 × 2 × 5)
Now, we can rewrite ^3√40 as a product of an integer and another irrational cube root:
^3√40 = ^3√(2 × 2 × 2 × 5) = ^3√(2^3 × 5) = 2 ^3√5
Therefore, the rewrite of ^3√40 as a product of an integer and another irrational cube root is 2 ^3√5.
The correct answer is option 1: 2 ^3√5.
The answer is option 3, 8 ^3√5.
To see why, we can simplify the expression ^3√40. The cube root of 40 can be written as 40^(1/3).
To rewrite 40 as a product of an integer and another cube root, we can factorize 40. We notice that 40 can be written as 8 * 5.
So, 40^(1/3) = (8 * 5)^(1/3) = 8^(1/3) * 5^(1/3) = 8 ^3√5.
Therefore, the expression ^3√40 can be rewritten as a product of an integer (8) and another irrational cube root ( ^3√5).