What is the irrational cube root root(81, 3) rewritten as a product of an integer and another irrational cube root?
Note: You will find the option to add a cubic root symbol in the Algebra (x) keyboard.
(1 point)
The irrational cube root of 81 can be written as 3∛(3).
The irrational cube root \(\sqrt[3]{81}\) can be rewritten as a product of an integer and another irrational cube root as follows:
\(\sqrt[3]{81} = \sqrt[3]{27 \times 3} = \sqrt[3]{27} \times \sqrt[3]{3}\)
The cube root of 27 can be simplified as \(3\), so the expression becomes:
\(\sqrt[3]{81} = 3\sqrt[3]{3}\)
To rewrite the cube root of 81 as a product of an integer and another irrational cube root, we can first simplify the cube root of 81.
The cube root of 81, denoted as ∛81, is the number that, when multiplied by itself three times, gives the value 81.
We can find this by finding the prime factorization of 81, which is 3 x 3 x 3 x 3.
Therefore, ∛81 = ∛(3 x 3 x 3 x 3).
Now, we can separate one cube root of 3 from the other three cube roots of 3:
∛81 = ∛(3 x 3 x 3 x 3) = 3∛3.
So, the irrational cube root of 81, ∛81, can be rewritten as the product of an integer (3) and another irrational cube root (∛3).