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)}