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

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

\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3zx\right]\)

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

\(=\frac{1}{2}\left(x+y+z\right)\left(x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\right)\)

\(=\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)

Suy ra \(\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{1}{2}\left(x+y+z\right)\)

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

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

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

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

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

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

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 

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

a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)

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

=a+b+c

b: 

Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

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

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

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

a) \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}\)

\(=a+b+c\)