What is the irrational cube root 81−−√3 rewritten as a product of an integer and another irrational cube root?
The cube root of 81 can be written as 3, since 3^3 = 81. So, we can rewrite ∛81 as ∛(3^3).
Therefore, the expression ∛81−−√ can be rewritten as 3 −−√ = ∛(3^3)−−√ = 3∛1 = 3(1) = 3.
Therefore, the irrational cube root ∛81−−√3 is rewritten as a product of an integer (3) and another irrational cube root (∛1).
To rewrite the irrational cube root ∛81 as a product of an integer and another irrational cube root, we can use the property that the cube root of a product is equal to the product of the cube roots of individual numbers.
So, let's simplify:
∛81 = ∛(3^4)
Using the property mentioned earlier, we can rewrite this as:
∛81 = ∛(3^3 * 3)
Now, we can separate the cube root of 3^3 and the cube root of 3:
∛81 = ∛(3^3) * ∛3
Simplifying further:
∛81 = 3∛3
Therefore, the irrational cube root ∛81 can be rewritten as a product of the integer 3 and the irrational cube root ∛3.
To rewrite the irrational cube root ∛81 as a product of an integer and another irrational cube root, we can follow these steps:
Step 1: Find the prime factors of the given number, 81.
81 = 3 × 3 × 3 × 3
Step 2: Group the prime factors in sets of three to form perfect cubes.
81 = (3 × 3 × 3) × (3 × 3 × 3)
Step 3: Take the cube root of each perfect cube.
∛81 = ∛(3 × 3 × 3) × (3 × 3 × 3)
Step 4: Simplify the expression.
∛81 = 3 × ∛(3 × 3 × 3)
Step 5: Rewrite the result as a product of an integer and another irrational cube root.
∛81 = 3∛27
Therefore, the irrational cube root ∛81 can be rewritten as a product of an integer and another irrational cube root: 3∛27.