$\\$
`x^3 + y^3 + z^3 = 3xyz`
`-> x^3 + y^3 +z^3 - 3xyz=0`
`-> (x+y)^3 - 3xy (x+y) + z^3 - 3xyz=0`
`-> (x+y+z)[(x+y)^2 - (x+y)z + z^2] - 3xy (x+y+z)=0`
`-> (x+y+z) (x^2 + 2xy + y^2 - xz - yz +z^2) - 3xy (x+y+z)=0`
`-> (x+y+z) (x^2 + y^2 +z^2 +2xy - xz - yz - 3xy)=0`
`-> (x+y+z)(x^2 + y^2+z^2 - xy - yz - xz)=0`
`-> 1/2 (x+y+z) (2x^2 +2y^2 +2z^2 - 2xy - 2yz - 2xz)=0`
`-> 1/2 (x+y+z)[(x^2 - 2xy + y^2)+(y^2 -2yz +z^2) + (x^2 - 2xz +z^2)]=0`
`-> 1/2 (x+y+z) [(x-y)^2 + (y-z)^2 + (x-z)^2]=0`
`->` \(\left[ \begin{array}{l}x+y+z=0 (1)\\(x-y)^2 +(y-z)^2 + (x-z)^2=0\end{array} \right.\)
$\bullet$ `(x-y)^2 + (y-z)^2 + (x-z)^2=0`
Với mọi `x,y,z` có : $\begin{cases} (x-y)^2≥0\\(y-z)^2 ≥0\\(x-z)^2≥0\end{cases}$
`-> (x-y)^2 + (y-z)^2 + (x-z)^2 ≥0∀x,y,z`
Dấu "`=`" xảy ra khi :
`↔` $\begin{cases} (x-y)^2=0\\(y-z)^2=0\\(x-z)^2=0 \end{cases}$
`↔` $\begin{cases} x-y=0\\y-z=0\\x-z=0 \end{cases}$
`↔` $\begin{cases} x=y\\y=z\\x=z \end{cases}$
`↔ x=y=z (2)`
Từ (1), (2)
`-> x=y=z, x+y+z=0`
