1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)

b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

a. \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)

b. \(=\left(1-x\right)^3\)

\(x^2\left(x-1\right)-\left(x-1\right)=\left(x^2-1\right)\left(x-1\right)\)

\(x^3-x^2-x+1\)

\(=x^2\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)^2\cdot\left(x+1\right)\)

\(x^3-x^2-x+1=\left(x^3-x^2\right)-\left(x-1\right)=x^2\left(x-1\right)-\left(x-1\right)=\left(x^2-1\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)^2\)

x3 – 2x2 + x

= x.x2 – x.2x + x (Xuất hiện nhân tử chung là x)

= x(x2 – 2x + 1) (Xuất hiện hằng đẳng thức (2))

= x(x – 1)2

\(x^3+2x^2+x\)

\(=x\left(x^2+2x+1\right)\)

\(=x\left(x+1\right)^2\)

\(x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)

\(\Leftrightarrow x^3-2x^2+x^2-2x+x-2\)

\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+1\right)\)

`x^3-x^2-x-2`

`=x^3-2x^2+x^2-2x+x-2`

`=(x^3-2x^2)+(x^2-2x)+(x-2)`

`=x^2(x-2)+x(x-2)+(x-2)`

`=(x-2)(x^2+x+1)`

`@` `\text {Ans}`

`\downarrow`

`x^3 - x^2 - x - 2`

`= x^3 - 2x^2 + x^2 - 2x + x - 2`

`= (x^3 - 2x^2) + (x^2 - 2x) + (x-2)`

`= x^2(x - 2) + x(x - 2) + (x-2)`

`= (x^2 + x + 1)(x-2)`

c: \(x^4+x^3-4x^2+x+1\)

\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)

\(=\left(x-1\right)\left(x^3+2x^2-2x-1\right)\)

\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\right]\)

\(=\left(x-1\right)^2\cdot\left(x^2+3x+1\right)\)

x3 – 2x2 + x – xy2

(Có nhân tử chung x)

= x(x2 – 2x + 1 – y2)

(Có x2 – 2x + 1 là hằng đẳng thức).

= x[(x – 1)2 – y2]

(Xuất hiện hằng đẳng thức (3))

= x(x – 1 + y)(x – 1 – y)

2(x+3)-x³-3x

\(=-x^3-3x+2x+6\)

\(=-x^3-x+6\)

Đa thức này ko phân tích được nha bạn

\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)

\(x^3-x+3x^2+3xy^2+y^3-y\)

\(=\left(x+y\right)^3-\left(x+y\right)\)

\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)

= (x³ + 3x²y + 3xy² + y³) - (x+y)

= (x + y)³ - (x + y)

= (x + y).[(x+y)² - 1 ]

= (x + y).(x + y - 1).(x + y + 1)

1: =(x-1-y)(x-1+y)

3: =(x-1)(x+1)(x-2)