What is the irrational cube root ^3√40 rewritten as a product of an integer and another irrational cube root?

The irrational cube root ^3√40 can be rewritten as a product of an integer and another irrational cube root by finding its prime factorization.

The prime factorization of 40 is 2^3 * 5.

Therefore, we can rewrite ^3√40 as ^3√(2^3 * 5).

Taking the cube root of 2^3 gives ∛(2^3) = 2.

The remaining factor is 5, which is a prime number and cannot be simplified further.

Therefore, ^3√40 can be rewritten as 2∛5, which is a product of an integer (2) and another irrational cube root (∛5).

To rewrite the irrational cube root ^3√40 as a product of an integer and another irrational cube root, we can start by finding the prime factorization of 40.

The prime factorization of 40 is: 2^3 * 5.

Since we are looking for an integer and another irrational cube root, we can rewrite 40 as a perfect cube multiplied by a different factor.

Let's rewrite 40 as (2^3) * (2 * 5).

Now, let's rewrite the cube root.

^3√(40) = ^3√[(2^3) * (2 * 5)]

Using the property of exponents, we can split the cube root of a product into a product of cube roots:

^3√[(2^3) * (2 * 5)] = ^3√(2^3) * ^3√(2 * 5)

Simplifying this expression, we get:

^3√(2^3) * ^3√(2 * 5) = 2 * ^3√(10)

Thus, the irrational cube root ^3√40 rewritten as a product of an integer and another irrational cube root is 2 * ^3√10.

To rewrite ^3√40 as a product of an integer and another irrational cube root, you need to find the largest possible perfect cube factor of 40. Here's how:

1. Start by finding the prime factorization of 40: 40 = 2^3 * 5.
2. Identify the largest perfect cube factor. In this case, it is 2^3 = 8, as 8 is a perfect cube.
3. Rewrite 40 as the product of the perfect cube factor and the remaining factor: 40 = 8 * 5.
4. Simplify and rearrange the expression: ^3√40 = ^3√(8 * 5) = ^3√8 * ^3√5.
5. Since ^3√8 = 2, the final simplified expression is: ^3√40 = 2 * ^3√5.

Therefore, the irrational cube root ^3√40 can be written as the product of the integer 2 and the irrational cube root ^3√5.