Đặt \(x-y=c,y-z=a,z-x=b\) thì \(a+b+c=0\Rightarrow c=-a+b\) , ta có:

\(P=\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)

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

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

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

\(P=3abc=3\left(x-y\right)\left(y-z\right)\left(z-x\right).\)

Hoặc cách khác:

Dễ dàng thấy x,y,z có thể hoán vị vòng quanh.Nên P chứa các nhân tử (x-y);(y-z);(z-x)

Sau khi phân tích P có dạng: \(k\left(x-y\right)\left(y-z\right)\left(z-x\right)\) (*)

Cho x = 1;y=2;z=3: \(P=\left(1-2\right)^3+\left(2-3\right)^3+\left(3-1\right)^3=6\)

Thay vào (*),ta có: \(P=k\left(x-y\right)\left(y-z\right)\left(z-x\right)=6\)

hay \(P=k\left(1-2\right)\left(2-3\right)\left(3-1\right)=6\)

Suy ra \(k=3\)

Thay k = 3 vào (*),ta phân tích được P thành nhân tử: \(P=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

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

Ta có:

 (x + y + z)² + (x + y – z)² – 4z²

^{\(=\left(x+y-z\right)^2+\left(x+y-z\right)\left(x+y+3z\right)\)}

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

\(\left(x+y+z\right)^2+\left(x+y-z\right)^2-4z^2\)

\(=x^2+y^2+z^2+2xy+2yz+2xz+x^2+y^2+z^2+2xy-2xz-2yz-4z^2\)

\(=2x^2+2y^2-2z^2+4xy\)

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

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

\(A=-x-z\left(x-y\right)+y=-x-xz+zy+y=-x\left(1+z\right)+y\left(1+z\right)=\left(1+z\right)\left(y-x\right)\)

A = -(x-y)-z(x-y)=(x-y)(-1-z)=(y-x)(z+1)

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

A = -x - zx + zy + y

A = -(-x - zx + zy + y)

A = x + zx - zy - y

A = x + zx - y - zy

A = x(1 + z) - y(1 + z)

A = (x - y)(1 + z)

Có sai đề không vậy bạn?

ko sai

a,Ta có: 
x³ + y³ + z³ - 3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz 
= [(x+y)³ + z³] - 3xy(x+y+z) 
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z) 
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy] 
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy) 
= (x+y+z)(x² + y² + z² - xy - xz - yz)

b, Từ: 
x + y + z = 0 
=> x + y = -z 
<=> (x + y)^3 = (-z)^3 
<=> x^3 + 3x^2y + 3xy^2 + y^3 = -z^3 
<=> x^3 + y^3 + z^3 = -3x^2y - 3xy^2 
<=> x^3 + y^3 + z^3 = -3xy(x+y) 
<=> x^3 + y^3 + z^3 = -3xy(-z) 
<=> x^3 + y^3 + z^3 = 3xyz 

bạn thu gom 2 đa thức đầu tiên thành 1 nhóm và 2 đa thức sau thành 1 nhóm . sau đó dùng hđt rồi đem chung 

nên nhớ 8=2³

\(4\left(x^2y^2+z^2t^2+2xyzt\right)-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left[2\left(xy+zt\right)\right]^2-\left(x^2+y^2-z^2-t^2\right)^2\)

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

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

Ta có: \(4\left(x^2y^2+2xyzt+z^2t^2\right)-\left(x^2+y^2-z^2-t^2\right)^2\)

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

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

\(=\left[-\left(x^2-2xy+y^2\right)+\left(z^2+2tz+t^2\right)\right]\left[\left(x^2+2xy+y^2\right)-\left(t^2-2tz+z^2\right)\right]\)

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

\(4(x^2y^2+z^2t^2+2xyzt)-(x^2+y^2-z^2-t^2)^2\)

\(=[2(xy+zt]^2-(x^2+y^2-z^2-t^2)^2\)

\(=(2xy+2zt)^2-(x^2+y^2-z^2-t^2)^2\)

\(=(2xy+2zt-x^2-y^2+z^2+t^2)(2xy+2zt+x^2+y^2-z^2-t^2)^2\)