a/ \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Vậy...
b/ \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2+z^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+\left[3\left(x+y\right)^2z+3\left(x+y\right)z^2\right]-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3\)
\(=3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)\)
\(=3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
Vậy..