(x+y)³-3x²y-3xy²+z³-3xyz 

=> ((x+y)³+z³)- 3xy(x+y+z)

=>(x+y+z)((x+y)²-z(x+y)+z²)-3xy(x+y+z)

=>(x+y+z)(x²+2xy+y²-xz-yz+z²-3xy)

vì x+y+z=0 => biểu thúc trên bằng 0

=> x³+y³+z³-3xyz=0

=>x³+y³+z³=3xyz

=>

=>

xét hiệu x³+y³+z³-3xyz

=(x+y)³+z³-3xy(x+y)-3xyz

=(x+y+z)³-3(x+y+z)(x+y)z-3xy(x+y+z)

=0       vì x+y+z=0

=>x³+y³+z³=3xyz

=>đpcm

Ta có; \(y^2=x\cdot z\)

=>\(\frac{x}{y}=\frac{y}{z}\) (1)

\(z^2=y\cdot t\)

=>\(\frac{y}{z}=\frac{z}{t}\) (2)

Từ (1),(2) suy ra \(\frac{x}{y}=\frac{y}{z}=\frac{z}{t}=k\)

=>\(\begin{cases}z=tk\\ y=zk=tk\cdot k=tk^2\\ x=y\cdot k=tk^2\cdot k=tk^3\end{cases}\)

\(\frac{x^3+y^3-2z^3}{y^3+z^3-2t^3}=\frac{\left(yk\right)^3+\left(zk\right)^3-2\cdot\left(tk\right)^3}{y^3+z^3-2t^3}=\frac{k^3\left(y^3+z^3-2t^3\right)}{y^3+z^3-2t^3}=k^3\)

\(\left(\frac{x+y-2z}{y+z-2t}\right)^3=\left(\frac{yk\cdot zk-2\cdot tk}{y+z-2t}\right)^3=k^3\)

Do đó: \(\frac{x^3+y^3-2z^3}{y^3+z^3-2t^3}=\left(\frac{x+y-2z}{y+z-2t}\right)^3\)

Ta có x+y+z = 0

\(\Rightarrow\left(+\right)x+y=-z\)

  \(\left(+\right)x+z=-y\)

\(\left(+\right)z+y=-x\)

C= (x+y)(y+z)(x+z)=(-z)(-x)(-y)=-1(xyz)=-1. 2= -2

=>\(\left(x-1\right)^2\)= 0 => x -1 =0 => x=1

=>\(2.|x-y|=0\)=> x -y = 0 => x=y =1

=>\(\left(x.y.z-3\right)^{2016}=0\)=>(x.y.z-3)=0

=> 1.1.z -3 =0 => z= 3

Vậy x=y=1, z=3