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

Sure, let's use De Moivre's theorem to find (√3 + i)^3.

De Moivre's theorem states that (cos θ + i sin θ)^n = cos(nθ) + i sin(nθ), where n is a real number.

First, we need to express √3 + i in polar form, r(cos θ + i sin θ).

The modulus r is given by √((√3)^2 + 1^2) = 2.

The argument θ is given by arctan(1/√3) = π/6.

So, √3 + i = 2(cos(π/6) + i sin(π/6)).

Now, we can apply De Moivre's theorem to find (√3 + i)^3.

(√3 + i)^3 = 2^3 [cos(3π/6) + i sin(3π/6)] = 8[cos(π/2) + i sin(π/2)] = 8i.

So, (√3 + i)^3 = 8i.