What is the irrational cube root 81−−√3 rewritten as a product of an integer and another irrational cube root?
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To rewrite the irrational cube root as a product of an integer and another irrational cube root, we need to find a perfect cube that can be factored out of 81.
81 can be expressed as (3^4) since 3^4 = 81.
81^(1/3) can be written as (3^4)^(1/3).
Using the property of exponents, we can rewrite it as:
(3^4)^(1/3) = 3^(4/3).
Therefore, the irrational cube root 81^(1/3) can be written as the product 3 * (3^(1/3)).
To rewrite the irrational cube root ∛(81-√3) as a product of an integer and another irrational cube root, we can simplify the expression using the property of cube roots.
Let's start by simplifying the cube root of 81:
∛(81) = 4
Next, let's simplify the cube root of √3:
∛(√3) cannot be simplified further, as √3 is an irrational number.
Now, we can rewrite the original expression as a product:
∛(81-√3) = ∛(81) * ∛(√3)
Plugging in the simplified expressions, we get:
∛(81-√3) = 4 * ∛(√3)
Therefore, the irrational cube root ∛(81-√3) can be rewritten as the product of the integer 4 and the irrational cube root ∛(√3).
To rewrite the irrational cube root 81√3 as a product of an integer and another irrational cube root, we need to find a perfect cube that is a multiple of 81.
81 is a perfect cube because it is equal to (3^4). Taking the cube root of 81, we get 3.
So, we can rewrite 81√3 as (3 * √3).
Therefore, the irrational cube root 81√3 can be rewritten as the product of 3 and the irrational cube root √3.