Question

因式分解:x2y+y2z+z2x+xy2+yz2+zx2+3xyz

Answer 1

原式=(x^2y+zx^2+xyz)+(y^2z+xy^2+xyz)+(z^2x+yz^2+xyz)
=x(xy+xz+yz)+y(yz+xy+xz)+z(xz+yz+xy)
=(xy+xz+yz)(x+y+z)

Answer 2

x2y+y2z+z2x+xy2+yz2+x2z+xyz+xyz+xyz=xy(x+y+z)+xz(x+y+z)+yz(x+y+z)=(x+y+z)(xy+xz+yz)