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\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)^3-3\left(x+y\right)z\left(x+y+z\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3zx\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(=\frac{1}{2}\left(x+y+z\right)\left(x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\right)\)
\(=\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)
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
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 = \(\frac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-xz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)\(=\frac{x-y+z}{2}\)
