a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)

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

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

a) Ta có: \(x^4+2x^3-4x-4\)

\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)

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

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

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

d) Ta có: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

t chỉ cho kết quả thôi nhá, còn nhóm nhân tử you tự xử nhá !

=(x-y)(z-x)(z-y)(x+y+z)

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

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

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

\(=xz^3-yz^3+y^3z-xy^3+\left(y-z\right)x^3\)

\(=xz^3-yz^3+y^3z-xy^3+y^3z-xy^3+x^3y-x^3z\)

Mk ko chắc

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

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

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

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

\(=\left(x-y\right)\left(z-y\right)\left(z^2+zy+y^2\right)+\left(y-z\right)\left(x-y\right)\left(x^2+y^2+xy\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(x^2+y^2+xy-z^2-y^2-zy\right)\)

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

Ta có: (x-y)^3+(y-z)^3+(z-x)^3 
Bạn để ý thấy (x-y)^3+(y-z)^3 là hằng đẳng thức dạng A^3+B^3=(A+B)(A^2-AB+B^2). Vậy ta có thể phân tích (x-y)^3+(y-z)^3 như sau 
(x-y+y-z)((x-y)^2-(x-y)(y-z)+(y-z)^2) 
(x-z)((x-y)^2-(x-y)(y-z)+(y-z)^2) 
-(z-x)((x-y)^2-(x-y)(y-z)+(y-z)^2) 

cách khác:

Đặt:   \(x-y=a;\)\(y-z=b;\)\(z-x=c\)

suy ra:    \(a+b+c=0\)

=>  \(a+b=-c\)

=>  \(\left(a+b\right)^3=-c^3\)

=>  \(a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3\)

<=>  \(a^3+b^3+c^3=-3ab\left(a+b\right)\)

<=>  \(a^3+b^3+c^3=-3ab\left(-c\right)=3abc\)

Thay trở lại đc:    \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

\(A=x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)=x\left(y^2-z^2\right)+y\left(-y^2+z^2-x^2+y^2\right)+z\left(x^2-y^2\right)=\left(y^2-z^2\right)\left(x-y\right)+\left(x^2-y^2\right)\left(z-y\right)=\left(y-z\right)\left(y+z\right)\left(x-y\right)-\left(x-y\right)\left(x+y\right)\left(y-z\right)=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c+abc+b^2c\right)\)

\(C=ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b-a+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b\right)+bc\left(a-c\right)+ac\left(a-c\right)=b\left(a+b\right)\left(a-c\right)+c\left(a-c\right)\left(a+b\right)=\left(a+b\right)\left(c+c\right)\left(a-c\right)\)