To find $(x^2 - \frac{1}{x^2})$ when $x = 3 + \sqrt{2}$, let's go through the steps.

Step 1: Calculate $x^2$

Given $x = 3 + \sqrt{2}$, we square both sides:

$x^2 = (3 + \sqrt{2})^2 = 3^2 + 2 \cdot 3 \cdot \sqrt{2} + (\sqrt{2})^2$

$x^2 = 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2}$

Step 2: Calculate $\frac{1}{x^2}$

Now, let's find $\frac{1}{x^2}$. First, recall that the conjugate of $x = 3 + \sqrt{2}$ is $3 - \sqrt{2}$, so:

$\frac{1}{x} = \frac{1}{3 + \sqrt{2}} \cdot \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{3 - \sqrt{2}}{(3 + \sqrt{2})(3 - \sqrt{2})} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}$

Thus,

$\frac{1}{x} = \frac{3 - \sqrt{2}}{7}$

To square this:

$\frac{1}{x^2} = \left(\frac{3 - \sqrt{2}}{7}\right)^2 = \frac{(3 - \sqrt{2})^2}{7^2} = \frac{9 - 6\sqrt{2} + 2}{49} = \frac{11 - 6\sqrt{2}}{49}$

Step 3: Calculate $x^2 - \frac{1}{x^2}$

Now subtract $\frac{1}{x^2}$ from $x^2$:

$x^2 - \frac{1}{x^2} = (11 + 6\sqrt{2}) - \frac{11 - 6\sqrt{2}}{49}$

We need to simplify this expression further. Let me handle the fraction carefully and compute the result.

The expression $x^2 - \frac{1}{x^2}$ simplifies to:

$x^2 - \frac{1}{x^2} = \frac{528}{49} + \frac{300\sqrt{2}}{49}$

If you'd like, I can provide further details or verify any specific steps.

Would you like more details or have any questions? Here are five related questions you might find helpful:

1. How do you calculate the conjugate of a binomial involving a square root?
2. Can you explain the process of rationalizing a denominator?
3. What is the significance of conjugates in simplifying expressions?
4. How would you apply this method to a different binomial like $5 + \sqrt{3}$?
5. How can you check if your result is correct after performing algebraic operations?

Tip: When simplifying algebraic expressions, especially those involving square roots, using conjugates is a useful tool to eliminate radicals from denominators.