What are the real and complex solutions of the polynomial equation x³ + 8 = 0?

Finding Solutions to the Polynomial Equation x³ + 8 = 0

The equation you provided is a cubic polynomial equation: x³ + 8 = 0.

To solve this, we can start by rewriting the equation:

x³ = -8

Next, we can take the cube root of both sides to find the real solution:

x = ∛(-8)

The cube root of -8 is -2, so we have:

x = -2

This gives us one real solution:

Real Solution: x = -2

Finding Complex Solutions

Now, to find the complex solutions, we can use the fact that the roots of the polynomial can be expressed in terms of the cube root of unity. The general formula for the roots of the equation x³ + a = 0 (where a = 8) can be given by:

x = ∛a * (1 + ω^k)

where:

ω = e^(2πi/3) is a cube root of unity,
k = 0, 1, 2 (this accounts for the three roots).

For our equation:

a = 8, so ∛8 = 2.

Now we calculate the roots:

1. First Root (k = 0):

x₁ = 2(1 + ω^0) = 2(1) = 2.

2. Second Root (k = 1):

x₂ = 2(1 + ω^1) = 2(1 + e^(2πi/3)).

Calculating ω^1, we find:

ω^1 = -1/2 + i(√3/2).

This gives:

x₂ = 2(1 - 1/2 + i(√3/2)) = 2(1/2 + i(√3/2)) = 1 + i√3.

3. Third Root (k = 2):

x₃ = 2(1 + ω^2) = 2(1 + e^(-2πi/3)).

Calculating ω^2, we get:

ω^2 = -1/2 - i(√3/2).