Ta có :
$x^3.(y-z)+z^3.(x-y)=y^3.(x-z)$
$ \to x^3.(y-z)+z^3.(x-y) + y^3.(z-x) = 0 $
$\to x^3.(y-z) + z^3x-z^3y-y^3z+y^3x=0$
$\to (y-z).(x^3-xy^2-xyz-xz^2+y^2z+yz^2)=0$
$\to (y-z).[x^2.(x-y)+x^2y-xy^2-xyz-xz^2+y^2z+z^2]=0$
$\to (y-z).[x^2.(x-y)-yz.(x-y)+xy.(x-y)-z^2.(x-y)]=0$
$\to (y-z).(x-y).(x^2-yz+xy-z^2)=0$
$\to (y-z).(x-y).[z.(x-z)-zx+x^2+y.(z-x)] = 0 $
$\to (y-z).(x-y).(x-z).(x+y+z)=0$
$\to$ $x+y+z=0$
$\to x+y = -z$
$\to x^3+y^3+3xy.(x+y)=-z^3$
$\to x^3+y^3+z^3 = 3xyz$ $(đpcm)$
