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The question asks to solve the equation \(\sqrt{x + \frac{1}{x}} = 6\) with \(x > 1\) and find the value of \(\left( x^2 - \frac{1}{x^2} \right)\).

Let's solve this step-by-step:

1. Start with the given equation:
   \[
   \sqrt{x + \frac{1}{x}} = 6
   \]
   Squaring both sides:
   \[
   x + \frac{1}{x} = 36
   \]

2. Now, to find the value of \( \left( x^2 - \frac{1}{x^2} \right) \), we can use the identity:
   \[
   \left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2
   \]
   Substituting \(x + \frac{1}{x} = 36\):
   \[
   36^2 = x^2 + \frac{1}{x^2} + 2
   \]
   \[
   1296 = x^2 + \frac{1}{x^2} + 2
   \]
   \[
   x^2 + \frac{1}{x^2} = 1296 - 2 = 1294
   \]

3. Now we need to find \( \left( x^2 - \frac{1}{x^2} \right) \). Using the identity:
   \[
   \left( x^2 - \frac{1}{x^2} \right) = \sqrt{ \left( x^2 + \frac{1}{x^2} \right)^2 - 4 }
   \]
   We already know \( x^2 + \frac{1}{x^2} = 1294 \), so:
   \[
   \left( x^2 - \frac{1}{x^2} \right) = \sqrt{1294^2 - 4}
   \]
   Simplifying this will give the final answer. Let me compute that.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like a detailed explanation of the steps so far, or would you prefer to retry later? Let me know if you need anything else.

Here are 5 related questions to explore further:
1. How do you find the square of binomials like \( \left(x + \frac{1}{x}\right) \)?
2. What are other methods to solve radical equations?
3. How does the identity for \(x^2 - \frac{1}{x^2}\) help in simplifications?
4. How can you use algebraic identities in different contexts?
5. What happens when squaring both sides of an equation introduces extraneous solutions?

**Tip**: Always check for extraneous solutions when squaring both sides of an equation.

If \(\sqrt{x + \frac{1}{x}} = 6\) and \(x > 1\), find the value of \(x^2 - \frac{1}{x^2}\).

Learn how to solve the radical equation \(\sqrt{x + \frac{1}{x}} = 6\) and find the value of \(x^2 - \frac{1}{x^2}\) step by step. This problem involves algebraic identities and is suitable for high school students in Grades 9-11.

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