Rewrite 3 square root of 40 as a product of integer and another irrational cube root
To rewrite 3√40 as a product of an integer and another irrational cube root, we need to find a cube root of 40 that when multiplied by an integer will give us the original value.
Let's find the prime factorization of 40: 40 = 2 * 2 * 2 * 5.
Now, we can rewrite 3√40 as follows:
3√40 = 3√(2 * 2 * 2 * 5)
To rewrite this expression, we can split it into two cube roots:
= 3√(2 * 2) * √(2 * 5)
= 3 * √(2^2) * √(10)
= 3 * 2 * √10
= 6√10
Therefore, 3√40 can be rewritten as the product of 6 and √10, where 6 is an integer and √10 is another irrational cube root.