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To rationalize the denominator of the given expression \(\frac{6}{3 - \sqrt{7}}\), you need to follow these steps:

### Step 1: Multiply by the Conjugate
The denominator is \(3 - \sqrt{7}\), so we multiply both the numerator and denominator by the conjugate of the denominator, which is \(3 + \sqrt{7}\). The purpose of this step is to eliminate the square root in the denominator.

\[
\frac{6}{3 - \sqrt{7}} \times \frac{3 + \sqrt{7}}{3 + \sqrt{7}} = \frac{6(3 + \sqrt{7})}{(3 - \sqrt{7})(3 + \sqrt{7})}
\]

### Step 2: Apply Difference of Squares in the Denominator
The denominator is now a difference of squares:

\[
(3 - \sqrt{7})(3 + \sqrt{7}) = 3^2 - (\sqrt{7})^2 = 9 - 7 = 2
\]

Thus, the denominator simplifies to 2.

### Step 3: Expand the Numerator
Now, expand the numerator:

\[
6(3 + \sqrt{7}) = 6 \times 3 + 6 \times \sqrt{7} = 18 + 6\sqrt{7}
\]

### Step 4: Simplify the Expression
Now, the expression becomes:

\[
\frac{18 + 6\sqrt{7}}{2}
\]

You can now simplify by dividing both terms in the numerator by 2:

\[
\frac{18}{2} + \frac{6\sqrt{7}}{2} = 9 + 3\sqrt{7}
\]

### Final Answer:
\[
9 + 3\sqrt{7}
\]

Would you like further details on any step?

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Here are 5 related questions that expand on this concept:
1. How would you rationalize a denominator like \(\frac{5}{2 + \sqrt{3}}\)?
2. Can you explain the difference of squares method in more detail?
3. How do you rationalize complex expressions with multiple radicals in the denominator?
4. What happens if both terms in the denominator contain square roots?
5. Could you rationalize a cube root in the denominator? If so, how?

**Tip:** When rationalizing a denominator, always look for opportunities to use the conjugate if there's a square root involved!

Without using a calculator, rationalise the denominator of \(\frac{6}{3 - \sqrt{7}}\). Simplify your answer. You must show each stage of your working.

Learn how to rationalise the denominator of \(\frac{6}{3 - \sqrt{7}}\) step-by-step using the conjugate method and the difference of squares theorem. Ideal for students in Grades 9-11.

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