X^3-2x+1=0

Найдите (2-a) (2-b) (2-c), где a, b, c корни кубического уравнения

X^3 - 2x + 1 = 0;

0=x^3 - x - x + 1 = x * (x^2 - 1) - (x-1) = x * (x-1) * (x+1) - (x-1) =

= (x-1) * (x * (x+1) - 1) = (x-1) * (x^2 + x - 1) = 0;

1) x-1 = 0, x_1=1

или

2) x^2 + x - 1 = 0;

D = 1 - 4 * (-1) = 1+4 = 5.

x_2 = (-1 - √5) / 2;

x_3 = (-1+√5) / 2;

(2-a) (2-b) (2-c) = (2 - x_1) * (2 - x_2) * (2 - x_3) =

= (2-1) * (2 - [ (-1-√5) / 2]) * (2 - [ (-1+√5) / 2]) =

= 1 * (4 + 1 + √5) * (4 + 1 - √5) / 4 = (5+√5) * (5-√5) / 4 = (25 - 5) / 4 = 20/4 = 5.