We are given:

$x + \frac{1}{x} = \sqrt{71}$

We need to find the value of $x^2 + \frac{1}{x^2}$.

Step-by-Step Process:

1. Square both sides of the equation to relate $x^2 + \frac{1}{x^2}$ to $x + \frac{1}{x}$.

   Squaring $x + \frac{1}{x}$:

   $\left( x + \frac{1}{x} \right)^2 = x^2 + 2 + \frac{1}{x^2}$

   Therefore:

   $x^2 + \frac{1}{x^2} + 2 = \left( \sqrt{71} \right)^2$

2. Simplify the equation:

   $x^2 + \frac{1}{x^2} + 2 = 71$

   Subtract 2 from both sides:

   $x^2 + \frac{1}{x^2} = 71 - 2$

   $x^2 + \frac{1}{x^2} = 69$

Thus, the value of $x^2 + \frac{1}{x^2}$ is $\boxed{69}$.

Would you like more details or have any questions?

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Here are 5 related questions for deeper understanding:

1. If $x + \frac{1}{x} = 5$, what is $x^2 + \frac{1}{x^2}$?
2. If $x + \frac{1}{x} = 10$, what is the value of $x^2 + \frac{1}{x^2}$?
3. How would you solve $x^2 + \frac{1}{x^2} = 45$ for $x + \frac{1}{x}$?
4. Can you express $x^3 + \frac{1}{x^3}$ in terms of $x + \frac{1}{x}$?
5. What would happen if $x + \frac{1}{x} = 0$? What is $x^2 + \frac{1}{x^2}$?

Tip: Always remember that squaring both sides of an equation can introduce extraneous solutions, so it's good to verify the solutions afterward if possible.