What is the irrational cube root ^3V81 rewritten as a product of an integer and another irrational cube root?
The irrational cube root of 81, ^3V81, can be rewritten as a product of an integer and another irrational cube root as follows:
\[ ^3V81 = ^3V(3^4) = ^3V(3^3 \times 3) = 3 \times \left(^3V3\right) \]
To rewrite the irrational cube root ^3V81 as a product of an integer and another irrational cube root, we need to find a perfect cube factor of 81.
The prime factorization of 81 is 3 * 3 * 3 * 3 = 3^4.
Taking the cube root of 81, we have:
^3V81 = ^3V(3^4) = 3^(4/3)
Therefore, the irrational cube root ^3V81 can be rewritten as the product of an integer and another irrational cube root:
^3V81 = 3^(4/3) = 3 * (^3V3^4) = 3 * (^3V27)
To rewrite the irrational cube root ^3√81 as a product of an integer and another irrational cube root, we need to find a perfect cube that can be factored out from 81.
To do this, we can consider the prime factorization of 81, which is 3 * 3 * 3 * 3 (or 3^4).
Now, let's rewrite ^3√81 as (^3√(3^4)).
Using the property of exponents, (^3√(3^4)) can be written as (^3√3^(4/3)).
We can simplify this further by writing 4/3 as 1 + 1/3.
Therefore, (^3√3^(4/3)) = (^3√3^(1+1/3)).
By using the property of exponents (a^(m+n) = a^m * a^n), this can be rewritten as (^3√3^1 * 3^(1/3)).
Simplifying further, (^3√3^1 * 3^(1/3)) is equal to (^3√3 * ^3√3^(1/3)).
Thus, the irrational cube root ^3√81 can be rewritten as a product of an integer (3) and another irrational cube root (^3√3^(1/3)).