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The equation you have provided asks to prove:

\[
(-1 + \sqrt{3}i)^{3n} + (-1 - \sqrt{3}i)^{3n} = 2^{3n+1}
\]

where \( n \) is any integer.

### Step-by-step solution:

1. **Recognize the terms**:
   - The complex numbers \( -1 + \sqrt{3}i \) and \( -1 - \sqrt{3}i \) are complex conjugates. These numbers can be written in polar form, which helps in understanding their powers.
   - These complex numbers resemble the cube roots of unity, specifically \( e^{2\pi i / 3} \) and \( e^{-2\pi i / 3} \). This is because \( -1 + \sqrt{3}i = 2e^{i2\pi/3} \) and \( -1 - \sqrt{3}i = 2e^{-i2\pi/3} \), with a modulus of 2 and argument \( \pm 120^\circ \) (or \( \pm \frac{2\pi}{3} \)).

2. **Use De Moivre's Theorem**:
   - De Moivre's Theorem states that for any complex number \( r(\cos \theta + i \sin \theta) \), its powers can be expressed as:
     \[
     (r (\cos \theta + i \sin \theta))^n = r^n (\cos(n\theta) + i \sin(n\theta))
     \]
   - Let’s apply this to the terms \( -1 + \sqrt{3}i \) and \( -1 - \sqrt{3}i \).

3. **Simplify both terms**:
   - First, consider \( (-1 + \sqrt{3}i) \) in polar form:
     \[
     -1 + \sqrt{3}i = 2\left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right)
     \]
     Similarly, \( (-1 - \sqrt{3}i) \) becomes:
     \[
     -1 - \sqrt{3}i = 2\left(\cos \frac{-2\pi}{3} + i \sin \frac{-2\pi}{3}\right)
     \]
   - Now raise both to the power of \( 3n \):
     \[
     (-1 + \sqrt{3}i)^{3n} = 2^{3n} \left( \cos(3n \cdot \frac{2\pi}{3}) + i \sin(3n \cdot \frac{2\pi}{3}) \right)
     \]
     \[
     (-1 - \sqrt{3}i)^{3n} = 2^{3n} \left( \cos(3n \cdot \frac{-2\pi}{3}) + i \sin(3n \cdot \frac{-2\pi}{3}) \right)
     \]

4. **Add the two terms**:
   - Since the sine terms will cancel each other out, you will only be left with the cosine terms:
     \[
     (-1 + \sqrt{3}i)^{3n} + (-1 - \sqrt{3}i)^{3n} = 2^{3n} \left( \cos(3n \cdot \frac{2\pi}{3}) + \cos(3n \cdot \frac{-2\pi}{3}) \right)
     \]
   - Using the fact that \( \cos \theta = \cos(-\theta) \), this simplifies to:
     \[
     = 2^{3n+1} \cos \left( 3n \cdot \frac{2\pi}{3} \right)
     \]

5. **Final cosine simplification**:
   - When \( 3n \) is an integer multiple of \( \frac{2\pi}{3} \), the cosine will cycle between 1 and -1, depending on \( n \). For integer values of \( n \), the cosine term becomes \( 1 \), leading to:
     \[
     (-1 + \sqrt{3}i)^{3n} + (-1 - \sqrt{3}i)^{3n} = 2^{3n+1}
     \]

Thus, we have proved that:

\[
(-1 + \sqrt{3}i)^{3n} + (-1 - \sqrt{3}i)^{3n} = 2^{3n+1}
\]

Let me know if you'd like further details or have questions!

### Related Questions:
1. How does De Moivre's theorem help with complex number powers?
2. What are cube roots of unity and how do they relate to this equation?
3. Can you express \( \cos(2\pi/3) \) and \( \sin(2\pi/3) \) geometrically?
4. How can polar form simplify calculations involving powers of complex numbers?
5. What is the role of the modulus in the polar representation of complex numbers?

### Tip:
Understanding complex numbers in both rectangular and polar forms is crucial when dealing with powers and roots. Always try converting between these forms for simplification.

Prove that (-1 + √3i)^(3n) + (-1 - √3i)^(3n) = 2^(3n+1), where n is any integer.

Learn how to prove the equation (-1 + √3i)^(3n) + (-1 - √3i)^(3n) = 2^(3n+1) using De Moivre's Theorem. This step-by-step solution is ideal for advanced high school students and covers essential concepts in complex numbers and polar forms.

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