(x+y)^3 - 3xy(x+y) + z^3 - 3xyz = 0
(x+y+z) ( (x+y)^2 +z^2 -z(x+y) -3xy) =0
(x+y+z) ( x^2+ 2xy+y^2 +z^2- zx-zy-3xy)=0
(x+y+z) ( x^2+y^2+z^2 -zx-zy -xy)=0
Suy ra x+y+z =0
x+y = -z
y+z = -x
x+z = -y
B = -16 + (-3) +2038 = 2019
Ta có: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\left(x,y,z\ne0\right)\)
+) x + y + z = 0 \(\Rightarrow B=\frac{-16z}{z}+\frac{-3x}{x}-\frac{-2038y}{y}\)
\(=-16-3+2038=2019\)
+) x = y = z \(\Rightarrow B=\frac{16.2z}{z}+\frac{3.2x}{x}-\frac{2038.2y}{y}\)
\(=32+6-4076=-4038\)
Cho em hỏi chút ạ, trường hợp x=y=z suy ra ntn ạ?
\(x^2+y^2+z^2=zy+yz+zx\)(chắc em hiểu chỗ này)
\(2x^2+2y^2+2z^2=2zy+2yz+2zx\)
\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\Leftrightarrow x=y=z\)
