To simplify the expression

$\frac{(6 - \sqrt{5})(6 + \sqrt{5})}{\sqrt{31}}$

we will proceed by using the difference of squares formula in the numerator.

1. Apply the Difference of Squares: Notice that $(6 - \sqrt{5})(6 + \sqrt{5})$ follows the form $(a - b)(a + b) = a^2 - b^2$.

   So, let: $a = 6 \quad \text{and} \quad b = \sqrt{5}$

   Then, $(6 - \sqrt{5})(6 + \sqrt{5}) = 6^2 - (\sqrt{5})^2 = 36 - 5 = 31$

2. Substitute Back into the Expression: Now, the expression simplifies to $\frac{31}{\sqrt{31}}$

3. Rationalize the Denominator: To further simplify, multiply both the numerator and the denominator by $\sqrt{31}$: $\frac{31 \cdot \sqrt{31}}{\sqrt{31} \cdot \sqrt{31}} = \frac{31\sqrt{31}}{31} = \sqrt{31}$

Thus, the simplified result is:

$\sqrt{31}$

Would you like a detailed breakdown of each step, or do you have any questions?

Here are some related questions to deepen understanding:

1. How would you simplify similar expressions involving differences of squares?
2. What is the process of rationalizing the denominator, and why is it used?
3. How can the difference of squares be applied to other algebraic expressions?
4. Can you give another example of rationalizing a denominator with a square root?
5. How does the concept of conjugates help in simplifying expressions?

Tip: The difference of squares formula $(a - b)(a + b) = a^2 - b^2$ is a powerful tool for simplifying expressions that involve products of terms with square roots.