Let's simplify the expression $(\sqrt{2} - \sqrt{3})(\sqrt{6} - \sqrt{2})$.

We will expand it using the distributive property (also known as FOIL for binomials):

$(\sqrt{2} - \sqrt{3})(\sqrt{6} - \sqrt{2}) = \sqrt{2} \cdot \sqrt{6} - \sqrt{2} \cdot \sqrt{2} - \sqrt{3} \cdot \sqrt{6} + \sqrt{3} \cdot \sqrt{2}$

Now, let's simplify each term:

1. $\sqrt{2} \cdot \sqrt{6} = \sqrt{12} = 2\sqrt{3}$
2. $\sqrt{2} \cdot \sqrt{2} = \sqrt{4} = 2$
3. $\sqrt{3} \cdot \sqrt{6} = \sqrt{18} = 3\sqrt{2}$
4. $\sqrt{3} \cdot \sqrt{2} = \sqrt{6}$

Now substitute these simplified terms back into the expression:

$2\sqrt{3} - 2 - 3\sqrt{2} + \sqrt{6}$

So the simplified expression is:

$2\sqrt{3} - 3\sqrt{2} + \sqrt{6} - 2$

This is the most simplified form of the expression.

Do you have any questions or need further explanation?

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Here are 5 related questions to explore:

1. How do you multiply two square root expressions in general?
2. How do you simplify expressions like $\sqrt{12}$ or $\sqrt{18}$?
3. What are the distributive and FOIL properties in algebra?
4. How can you add or subtract like terms when square roots are involved?
5. What happens when you multiply two conjugate binomials like $(a + b)(a - b)$?

Tip: When simplifying square roots, look for perfect squares within the radicand to make the process easier!