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?