Rewrite the irrational cube root 189−−−√3

Question

as the product of an integer and another irrational cube root. Show your work.

Note: You will find the option to add a cubic root symbol in the Algebra ( × )

Answer 1

To rewrite \( \sqrt[3]{189} \) as the product of an integer and another irrational cube root, we can start by factoring 189 into its prime factors.

First, let's find the prime factorization of 189:

1. Divide by 3 (the smallest prime number): \[ 189 \div 3 = 63 \]

2. Divide by 3 again: \[ 63 \div 3 = 21 \]

3. Divide by 3 again: \[ 21 \div 3 = 7 \]

So, we have the prime factorization: \[ 189 = 3^3 \times 7 \]

Now, we express \( \sqrt[3]{189} \): \[ \sqrt[3]{189} = \sqrt[3]{3^3 \times 7} \]

Using the properties of cube roots, we can separate the cube root of the product: \[ \sqrt[3]{189} = \sqrt[3]{3^3} \times \sqrt[3]{7} \]

We know that \( \sqrt[3]{3^3} = 3 \). Therefore: \[ \sqrt[3]{189} = 3 \times \sqrt[3]{7} \]

Thus, we have successfully rewritten \( \sqrt[3]{189} \) as the product of an integer and another irrational cube root: \[ \sqrt[3]{189} = 3 \sqrt[3]{7} \]