Bonjour,

(x-2)(x-2)(x+1)= x³-3x²+4

(x²-2x-2x+4)(x+1)= x³-3x²+4

(x²-4x+4)(x+1)= x³-3x²+4

x³-4x²+4x+x²-4x+4= x³-3x²+4

x³-3x²+4 = x³-3x²+4

Résoudre :

x³= 3x²-4

x³-3x+4= 0

(x³+x²) -4x²+4= 0

x²(x+1) -4(x²-1)= 0 ***x²-1: IR

x²(x+1) -4(x-1)(x+1)= 0 *** on factorise

(x+1)(x²-4(x-1))= 0

(x+1)(x²-4x+4)= 0 *** on factorise

(x+1)(x-2)²= 0

on résout:

x+1= 0 ou x-2= 0

x= -1 x= 2

S= { - 1 ; 2 }