Răspuns:

am atașat rezolvarea

a^3+b^3=(a+b)(a^2-ab+b^2)

[tex]\it x^3-2x^2-2x+1=0 \Rightarrow x^3+x^2-3x^2-3x+x+1=0 \Rightarrow \\ \\ \Rightarrow x^2(x+1)-3x(x+1)+(x+1)=0 \Rightarrow (x+1)(x^2-3x+1)=0 \Rightarrow \\ \\ \Rightarrow \begin{cases}\it x+1=0 \Rightarrow x=-1\\ \\ sau\\ \\ \it x^2-3x+1=0 \Rightarrow x=\dfrac{3\pm\sqrt5}{2}\end{cases}\\ \\ \\ R\breve ad\breve a cinile\ polinomului\ sunt:\\ \\ x_1=-1,\ \ \ x_2=\dfrac{3-\sqrt5}{2}, \ \ \ x_3=\dfrac{3+\sqrt5}{2}[/tex]