Ta có :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y+z\right)^3-x^3\right]-\left(y^3+z^3\right)\)
\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+x^2+\left(x+y+z\right)x\right]-\left(y+z\right)\left(y^2+z^2-yz\right)\)
\(=\left(y+z\right)\left[x^2+y^2+z^2+2\left(xy+yz+xz\right)+x^2+x^2+xy+zx\right]\)\(-\left(y+z\right)\left(y^2+z^2-yz\right)\)
\(=\left(y+z\right)\left[3x^2+y^2+z^2+3xy+3zx+2yz-y^2-z^2+yz\right]\)
\(=\left(y+z\right)\left[3x^2+3xy+3zx+3yz\right]\)
\(=\left(y+z\right)3\left[\left(x^2+xy\right)+\left(zx+yz\right)\right]\)
\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
Vậy ...
Ta có
(x+y+z)3-x3-y3-z3
=x3+y3+z3+3(x+y)(x+z)(z+y)-(x3+y3+z3)
=3(x+y)(x+z)(z+y)
