Math Olympiad Question | A Nice Algebra Problem | What Is The Value Of "X" In This Equation?

[@stpat7614]

x^3 = 6^3
Let a = x, and b = 6
x^3 = 6^3
=> a^3 = b^3
=> a^3 - b^3 = b^3 - b^3
=> a^3 - b^3 = 0
=> (a - b)(a^2 + a * b + b^2) = 0
=> (a - b)(a^2 + b * a + b^2) = 0
Suppose a - b = 0
Remember, a = x, and b = 6
a - b = 0
=> x - 6 = 0
=> x - 6 + 6 = 0 + 6
=> x = 6
x1 = 6
Suppose a^2 + b * a + b^2 = 0
1 * a^2 + b * a + b^2 = 0
a = (-b +/- sqrt[b^2 - 4 * 1 * b^2]) / (2 * 1)
a = (-b +/- sqrt[b^2 * 1 - b^2 * 4]) / (2)
a = (-b +/- sqrt[b^2 * (1 - 4)]) / 2
a = (-b +/- sqrt[b^2 * (-3)]) / 2
a = (-b +/- sqrt[b^2 * 3 * (-1)]) / 2
a = (-b +/- sqrt[b^2] * sqrt[3] * sqrt[-1]) / 2
a = (-b +/- b * sqrt[3] * i) / 2
a = b * (-1 +/- 1 * sqrt[3] * i) / 2
a = b * (-1 +/- sqrt[3] * i) / 2
Remember, a = x, and b = 6
a = b * (-1 +/- sqrt[3] * i) / 2
=> x = 6 * (-1 +/- sqrt[3] * i) / 2
=> x = ([1/2] * 6) * (-1 +/- sqrt[3] * i)
=> x = 3 * (-1 +/- sqrt[3] * i)
x = 3 * (-1 + sqrt[3] * i), or x = 3 * (-1 - sqrt[3] * i)
x = -3 * (1 - sqrt[3] * i), or x = -3 * (1 + sqrt[3] * i)
x2 = -3 * (1 - sqrt[3] * i)
x3 = -3 * (1 + sqrt[3] * i)
{x1, x2, x3} = {6, -3 * (1 - sqrt[3] * i), -3 * (1 + sqrt[3] * i)}

[@mukundchaffy3176]

Really?