Rewrite the irrational cube root 189−−−√3

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

1 answer

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

First, we can find the prime factorization of \(189\):

\[ 189 = 3 \times 63 = 3 \times 3 \times 21 = 3^3 \times 7 \]

Thus, we can express \(189\) as:

\[ 189 = 3^3 \times 7 \]

Now, we can rewrite \(\sqrt[3]{189}\):

\[ \sqrt[3]{189} = \sqrt[3]{3^3 \times 7} \]

Using the property of cube roots that states \(\sqrt[3]{a \times b} = \sqrt[3]{a} \cdot \sqrt[3]{b}\), we can split the expression:

\[ \sqrt[3]{189} = \sqrt[3]{3^3} \cdot \sqrt[3]{7} \]

Calculating \(\sqrt[3]{3^3}\), we find:

\[ \sqrt[3]{3^3} = 3 \]

Thus, the expression simplifies to:

\[ \sqrt[3]{189} = 3 \cdot \sqrt[3]{7} \]

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

\[ \sqrt[3]{189} = 3\sqrt[3]{7} \]