Ta có :
$(x+y+z)^3 = (x+y)^3 + 3(x+y)z(x+y+z) + z^3$
$= x^3+y^3+z^3+3xy(x+y)+3(x+y)(xz+yz+z^2)= x^3+y^3+z^3 + 3(x+y)(y+z)(x+z)$
Suy ra :
$x^3+y^3+z^3 - 3xyz = (x+y+z)^3 - 3(x+y)(y+z)(x+z) - 3xyz$
mà $(x+y)(y+z)(x+z) - xyz = (x+y+z)(xy+yz+zx)$
nên $x^3+y^3+z^3-3xyz = (x+y+z)^3 - 3(x+y+z)(xy+yz+xz) = (x+y+z)(x^2+y^2+z^2 - xy-yz-xz)$
- tranductucr1 và Kagome thích