Question

rationalize the denominator for

\\\frac{1}{\sqrt3{18}}\\

Explanation:

🆕 New Concept Discovered: Rationalizing Higher-Order Radicals
Multiply to complete perfect powers in the root.

Step 1: Analyze the denominator

The given expression is:

$$ \frac{1}{\sqrt[3]{18}} $$

First, find the prime factorization of the number inside the cube root:

$$ 18 = 2 \times 3^2 $$

So, the denominator can be written as:

$$ \sqrt[3]{2^1 \times 3^2} $$

Step 2: Determine the rationalizing factor

To eliminate the cube root in the denominator, we need the powers of all prime factors inside the root to be multiples of 3.

• For \( 2^1 \), we need to multiply by \( 2^2 \) to get \( 2^3 \).
• For \( 3^2 \), we need to multiply by \( 3^1 \) to get \( 3^3 \).

Therefore, the factor we need to multiply by inside a cube root is:

$$ \sqrt[3]{2^2 \times 3^1} = \sqrt[3]{4 \times 3} = \sqrt[3]{12} $$

Step 3: Multiply numerator and denominator

Multiply both the top and the bottom of the fraction by \( \sqrt[3]{12} \):

$$ \frac{1}{\sqrt[3]{18}} \times \frac{\sqrt[3]{12}}{\sqrt[3]{12}} = \frac{\sqrt[3]{12}}{\sqrt[3]{18 \times 12}} $$

Step 4: Simplify the denominator

Simplify the product inside the cube root in the denominator:

$$ 18 \times 12 = 216 $$

Since \( 216 = 6^3 \), we have:

$$ \sqrt[3]{216} = 6 $$

Putting it all together:

$$ \frac{\sqrt[3]{12}}{6} $$

Answer:

$$ \frac{\sqrt[3]{12}}{6} $$