If $x = 4 + \sqrt{6}$ and $y = 4 - \sqrt{6}$, then find the value of $x^2 + y^2$.

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**Solution:**

1. Given:
   - $x = 4 + \sqrt{6}$
   - $y = 4 - \sqrt{6}$

2. Use the identity:
   $$x^2 + y^2 = (x + y)^2 - 2xy$$

3. Calculate $x + y$:
   $$x + y = (4 + \sqrt{6}) + (4 - \sqrt{6}) = 8$$

4. Calculate $xy$:
   $$xy = (4 + \sqrt{6})(4 - \sqrt{6}) = 4^2 - (\sqrt{6})^2 = 16 - 6 = 10$$

5. Substitute values:
   $$x^2 + y^2 = (8)^2 - 2 	imes 10 = 64 - 20 = 44$$

**Answer:** $44$

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