(x+y)(y+z)(x+z)=8xyz
<=>\((xy+xz+y^2+yz)(x+z)=8xyz\)
<=>\(x^2y+x^2z+y^2z+xyz+xyz+xz^2+z^2y+yz^2=8xyz\)
<=> \(x^2y+x^2z+y^2x+xz^2+y^2z+yz^2-6xyz=0\)
<=> \(y(x^2+z^2-2xz)+x(y^2-2yz+z^2)+z(y^2-2yx+x^2)=0\)
<=>\(y(x-z)^2+x(y-z)^2+z(x-y)^2=0\)
Mà x,y,z dương
=> \((x-z)^2=0=>x=z\)
\((x-y)^2=0=>x=y\)
\((y-z)^2=0=>y=z\)
Vậy x=y=z