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

\(=\frac{\left(x-y\right)^3+z^3+3x^2y-3xy^2+3xyz}{2x^2+2y^2+2z^2+2xy+2yz-2xz}\)

\(=\frac{\left(x-y+z\right)\left[\left(x-y\right)^2+z^2-\left(x-y\right)z\right]+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\frac{\left(x-y+z\right)\left[x^2+y^2-2xy+z^2-xz+yz+3xy\right]}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\frac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\frac{x-y+z}{2}\)

x³−y³+z³+3xzy(x+y)²+(y+z)²+(z−x)² 

=(x−y)³+z³+3x²y−3xy²+3xyz2x²+2y²+2z²+2xy+2yz−2xz 

=(x−y+z)[(x−y)²+z²−(x−y)z]+3xy(x−y+z)2(x²+y²+z²+xy+yz−xz) 

=(x−y+z)[x²+y²−2xy+z²−xz+yz+3xy]2(x²+y²+z²+xy+yz−xz) 

=(x−y+z)(x²+y²+z²+xy+yz−xz)2(x²+y²+z²+xy+yz−xz) 

=x−y+z2 

\(=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}=\frac{x+y+z}{2}\)

Biến đổi tử thức ta đc:

   x³ - y³ + z³ + 3xyz

= (x - y)³ + z³ + 3x²y - 3xy² + 3xyz

= (x - y + z) [ (x - y)² - (x - y)z + z² ]  +  3xy(x - y + z)

= (x - y + z)(x² - 2xy + y² - xz + yz + z² + 3xy)

= (x - y + z)(x² + y² + z² + xy + yz - xz)

Biến đổi mẫu thức ta đc:

     (x + y)² + (y + z)² + (z - x)²

= x² + 2xy + y² + y² + 2yz + z² + z² - 2xz + x²

= 2(x² + y² + z² + xy + yz - xz)

 Vậy  A = \(\frac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)\(=\frac{x-y+z}{2}\)