Ask a New Question
Menu

Question

Rewrite the irrational cube root 189−−−√3 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 ( ×
) keyboard.

(2 points)

1 answer

To rewrite \( \sqrt[3]{189} \) as the product of an integer and another irrational cube root, we need to factor \( 189 \) into its prime factors and look for perfect cubes.

  1. Factor \( 189 \):

    • \( 189 \) can be factored as \( 3 \times 63 \)
    • \( 63 \) can be factored as \( 3 \times 21 \)
    • \( 21 \) can be factored as \( 3 \times 7 \)

    Therefore, \( 189 = 3^3 \times 7 \).

  2. Take the cube root:

    We can write: \[ \sqrt[3]{189} = \sqrt[3]{3^3 \times 7} \]

    By the property of cube roots that states \( \sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b} \), we can separate the cube root:

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

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

  3. Final expression:

    Thus, we can express \( \sqrt[3]{189} \) as: \[ \sqrt[3]{189} = 3 \sqrt[3]{7} \]

So the final answer is: \[ 3 \sqrt[3]{7} \]