To express $(\sqrt{3} + i)^6$ in the form $a + bi$, we can approach the solution by first converting $\sqrt{3} + i$ into polar form and then applying De Moivre's Theorem.

Step-by-Step Solution:

1. Convert $\sqrt{3} + i$ to Polar Form:

   • Let $z = \sqrt{3} + i$.
   • Calculate the modulus $|z|$: $|z| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$
   • Calculate the argument $\theta$: $\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$
   • Therefore, in polar form, $z = 2\left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)$.

2. Apply De Moivre's Theorem:

   • To find $z^6$, we use: $z^6 = \left(2 \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)\right)^6 = 2^6 \left(\cos \frac{6\pi}{6} + i \sin \frac{6\pi}{6}\right)$
   • Simplify: $2^6 = 64$ and $\frac{6\pi}{6} = \pi$ So, $z^6 = 64 \left(\cos \pi + i \sin \pi\right)$

3. Evaluate $\cos \pi$ and $\sin \pi$:

   • $\cos \pi = -1$ and $\sin \pi = 0$,
   • Thus, $z^6 = 64 \cdot (-1 + 0 \cdot i) = -64$

Final Answer:

$z^6 = -64$

Would you like further details or explanations on this problem?

Here are five questions that relate to this problem for further exploration:

1. How would you convert a complex number to polar form if it were located in a different quadrant?
2. What is De Moivre's Theorem, and how is it generally applied to powers of complex numbers?
3. Why is it necessary to convert the complex number to polar form before raising it to a power?
4. What happens if the exponent in De Moivre’s Theorem is a fraction instead of an integer?
5. Can you use De Moivre’s Theorem for negative exponents, and if so, how?

Tip: When working with powers of complex numbers, converting to polar form simplifies calculations significantly.