Решите уравнение:x^3-2x^2+x=(x^2-2x+1)^2

$x^3 - 2x^2 + x = (x^2 -2x + 1)^2\\\\ x(x^2 - 2x + 1) - (x^2 -2x + 1)^2 = 0\\\\ (x^2 - 2x + 1)(x - (x^2 - 2x + 1)) = 0\\\\ (x^2 - 2x + 1)(-x^2 + 3x - 1) = 0\\\\ (x^2 - 2x + 1)(x^2 - 3x + 1) = 0\\\\1) \ x^2 - 2x + 1 = x^2 - x - x + 1 =\\\\= x(x - 1) - (x - 1) = (x - 1)(x - 1) = (x - 1)^2 = 0$

$x_1 = x_2 = 1\\\\ 2) \ x^2 - 3x + 1 = 0\\\\ D = 9 - 4 = 5\\\\ x_1 = \frac{3 + \sqrt{5}}{2}\\\\ x_2 = \frac{3 - \sqrt{5}}{2}\\\\ \mathbb{OTBET:} \ x_1 = 1, \ x_2 = \frac{3 + \sqrt{5}}{2}, \ x_3 = \frac{3 - \sqrt{5}}{2}$