\(\left(x+1\right)^3-\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)\\ =\left[\left(x+1\right)-\left(x-1\right)\right].\left[\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]-\left(x^3-1\right)\\ =2.\left[\left(x^2+2x+1\right)+\left(x^2-1\right)+\left(x^2-2x+1\right)\right]-\left(x^3-1\right)\\ =2.\left(3x^2+1\right)-\left(x^3-1\right)\\ =6x^2+2-x^3+1=-x^3+6x^2+3\)
Bạn POP POP ghi thiếu (x^3-1) kìa bạn sửa giúp mik là làm lại ạ
\(\left(x+1\right)^3-\left(x-1\right)^3-\left(x^3-1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+3x^2+3x+1-x^3+3x^2-3x+1-x^3+1-\left(x^3-1^3\right)\\ =x^3+3x^2+3x+1-x^3+3x^2-3x+1-x^3+1-x^3+1\\ =-2x^3+6x^2+4\)