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

x²y+xyz+zx²+xy²+y²z+xyz+xyz+yz²+z²x=xyz

(x²y+xy²)+(xyz+zx²)+(y²z+xyz)+(yz²+z²x)+xyz=xyz

xy(x+y)+zx(y+x)+yz(y+x)+z²(y+x)+xyz=xyz

(x+y)(xy+xz+yz+z²)+xyz=xyz

(x+y)[(xy+xz)+(yz+z²)]+xyz=xyz

(x+y)[x(y+z)+z(y+z)]+xyz=xyz

(x+y)(x+z)(y+z)+xyz=xyz

(x+y)(x+z)(y+z)=xyz-xyz

(x+y)(x+z)(y+z)=0

=>\(\left[{}\begin{matrix}x+y=0\\x+z=0\\y+z=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-y\\x=-z\\y=-z\end{matrix}\right.\)

Với x=-z

=>VT= x²⁰¹⁵+y²⁰¹⁵+z²⁰¹⁵=(-z)²⁰¹⁵+z²⁰¹⁵+y²⁰¹⁵=y²⁰¹⁵

VP=(x+y+z)²⁰¹⁵=(-z+y+z)²⁰¹⁵=y²⁰¹⁵

Vậy x²⁰¹⁵+y²⁰¹⁵+z²⁰¹⁵=(x+y+z)²⁰¹⁵ với (x+y+z)(xy+yz+zx)=xyz

Có: \(x^2+y^2+z^2=xy+yz+xz\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2xz\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

\(\Leftrightarrow\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\)\(\Leftrightarrow x=y=z\)

Lại có: \(x^{2015}+y^{2015}+z^{2015}=3^{2016}\)

\(\Leftrightarrow x^{2015}+x^{2015}+x^{2015}=3^{2016}\)

\(\Leftrightarrow3x^{2015}=3^{2016}\)

\(\Leftrightarrow x=3\)

Vậy \(x=y=z=3\)