Use De Moivre’s theorem to find (√3 + i) 3 .

De Moivre's theorem states that for any complex number \( z = r(\cos \theta + i \sin \theta) \), where \( r \) is the modulus of \( z \) and \( \theta \) is the argument of \( z \), the \( n \)-th power of \( z \) is given by:

\[ z^n = r^n (\cos (n\theta) + i \sin (n\theta)) \]

For the complex number \( \sqrt{3} + i \), we can express it in polar form:

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

Now, raise \( \sqrt{3} + i \) to the power of 3 using De Moivre's theorem:

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

Simplify the expression:

\[ (\sqrt{3} + i)^3 = 8 \left(\cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right)\right) \]

Since \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \sin\left(\frac{\pi}{2}\right) = 1 \), the expression becomes:

\[ (\sqrt{3} + i)^3 = 8 \cdot (0 + i) = 8i \]

Therefore, \( (\sqrt{3} + i)^3 = 8i \).