x^3-2x+1=0Найдите (2-a)(2-b)(2-c), где a,b,c корешки 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, lt;=gt; 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.