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The image presents the following mathematical statement:

**If \( n \) is any integer, show that:**

\[
(\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^n \cos\left(\frac{n\pi}{6}\right)
\]

This is a complex number problem, where \( i \) is the imaginary unit. The goal is to prove that the sum of the powers of the two complex numbers \( (\sqrt{3} + i)^n \) and \( (\sqrt{3} - i)^n \) can be expressed in terms of \( 2^n \) and a cosine function.

### Approach to solve:

1. **Use Euler's formula:** 
   Convert the complex numbers \( \sqrt{3} + i \) and \( \sqrt{3} - i \) into polar form:
   - \( \sqrt{3} + i = 2 \text{cis} \left( \frac{\pi}{6} \right) \)
   - \( \sqrt{3} - i = 2 \text{cis} \left( -\frac{\pi}{6} \right) \)

   Here, \( \text{cis}(\theta) \) stands for \( \cos(\theta) + i\sin(\theta) \).

2. **Raise both to the power \( n \):**
   - \( (\sqrt{3} + i)^n = 2^n \text{cis} \left( \frac{n\pi}{6} \right) \)
   - \( (\sqrt{3} - i)^n = 2^n \text{cis} \left( -\frac{n\pi}{6} \right) \)

3. **Add the two results:**
   \[
   (\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^n \left[ \text{cis} \left( \frac{n\pi}{6} \right) + \text{cis} \left( -\frac{n\pi}{6} \right) \right]
   \]

4. **Simplify using the identity:**
   \[
   \text{cis}(\theta) + \text{cis}(-\theta) = 2\cos(\theta)
   \]
   Applying this identity:
   \[
   (\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^n \cdot 2 \cos \left( \frac{n\pi}{6} \right)
   \]

5. **Final expression:**
   \[
   (\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n+1} \cos \left( \frac{n\pi}{6} \right)
   \]
   The result matches the given expression, completing the proof.

Let me know if you need a more detailed explanation of any of the steps, or if you have further questions!

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Here are five related questions to expand on this topic:
1. How do you convert a complex number into polar form?
2. What is Euler's formula, and how is it useful in working with complex numbers?
3. Can you apply De Moivre's Theorem to similar problems involving trigonometric functions?
4. What is the geometric interpretation of adding two complex conjugates raised to a power?
5. How can you generalize this approach for other powers and angles?

**Tip:** Whenever you deal with complex numbers in power form, always consider expressing them in polar form—it simplifies the calculation significantly!

If n is any integer, show that (√3 + i)^n + (√3 - i)^n = 2^n * cos(nπ/6)

This problem involves complex numbers and trigonometry. It demonstrates the use of De Moivre's theorem and Euler's formula to simplify complex numbers raised to a power. Learn how to prove the identity (√3 + i)^n + (√3 - i)^n = 2^n * cos(nπ/6). Suitable for advanced high school or undergraduate math students.

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