Rewrite the irrational cube root 189−−−√3 as the product of an integer and another irrational cube root. Show your work.

Answer 1

We can rewrite the irrational cube root $\sqrt[3]{189}$ as the product of an integer and another irrational cube root by finding the perfect cube that is a factor of 189.

Finding the prime factorization of 189, we get:

$189 = 3 \times 3 \times 3 \times 7$

From this, we see that $3 \times 3 \times 3 = 27$ is a perfect cube that is a factor of 189.

So, we can rewrite $\sqrt[3]{189}$ as:

$\sqrt[3]{189} = \sqrt[3]{27 \times 7} = \sqrt[3]{27} \times \sqrt[3]{7} = 3\sqrt[3]{7}$

Therefore, $\sqrt[3]{189}$ can be written as the product of an integer ($3$) and another irrational cube root ($\sqrt[3]{7}$).