Solve the polynomial equation x^3-27=0 by factoring. (with explanations of each step)

x³ - 27 = 0

x³ + 3x² - 3x² + 9x - 9x - 27 = 0

x³ + 3x² + 9x - 3x² - 9x - 27 = 0

x (x²) + x (3x) + x (9) - 3 (x²) - 3 (3x) - 3 (9) = 0

x (x² + 3x + 9) - 3 (x² + 3x + 9) = 0

(x - 3) (x² + 3x + 9) = 0

x - 3 = 0 or x² + 3x + 9 = 0

+ 3 + 3 x = - (3) ± √ ((3) ² - 4 (1) (9))

x = 3 2 (1)

x = - 3 ± √ (9 - 36)

2

x = - 3 ± √ (-27)

2

x = - 3 ± 3i√ (3)

2

x = - 1.5 ± 1.5i√ (3)

x = - 1.5 + 1.5i√ (3) or x = - 1.5 - 1.5i√ (3)

In each step I have to factor the equation to a trinomial so that it could be factored into the sums and differences of cubes. Then, I have to use the quadratic equation in order to find the two solutions of the second factor, which is equal to - 1.5 + 1.5i√ (3) and - 1.5 - 1.5i√ (3), while on the first factor I have to find the solution to the first factor, making the solution equal to 3. So the solutions to the polynomial equation are the numbers 3, - 1.5 + 1.5i√ (3), and - 1.5 - 1.5i√ (3).

Solution Set: {3, - 1.5 ± 1.5i√ (3) }