[BLACK-Dragon]

ข้อ 1.4

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

$=(x+y+z)(xy+yz+zx)-xyz$

จากข้อ 1.5

$(y+z)(z+x)(x+y)=(x+y+z)(xy+yz+zx)-xyz$