Question

[^($_DUY_$)^]

Cho \(x^3+y^3+z^3=3xyz\) là \(x+y+z\ne0\)
C/M: \(a=b=c\)

Answer 1

[Nguyễn Lê Phước Thịnh]

Sửa đề: Chứng minh x=y=z

\(x^3+y^3+z^3=3xyz\)

=>\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

=>\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

=>\(\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz-3xy\right)=0\)

=>\(x^2+y^2+z^2-xy-xz-yz=0\)

=>\(2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

=>\(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)

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

=>x=y=z

Answer 2

[Toru]

Có:

\(x^3+y^3+z^3=3xyz\\\Leftrightarrow x^3+y^3+z^3-3xyz=0\\\Leftrightarrow(x+y)^3+z^3-3xy(x+y)-3xyz=0\\\Leftrightarrow (x+y+z)^3-3(x+y)z(x+y+z)-3xy(x+y+z)=0\\\Leftrightarrow (x+y+z)[(x+y+z)^2-3(x+y)z-3xy]=0\\\Leftrightarrow (x+y+z)(x^2+y^2+z^2+2xy+2yz+2xz-3xz-3yz-3xy)=0\\\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0\\\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0 (vì.x+y+z\neq0)\\\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\\\Leftrightarrow(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)=0\\\Leftrightarrow(x-y)^2+(y-z)^2+(x-z)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x;y\\\left(y-z\right)^2\ge0\forall x;y\\\left(x-z\right)^2\ge0\forall x;y\end{matrix}\right.\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x;y;z\)

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

nên: \(\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\x=z\end{matrix}\right.\Leftrightarrow x=y=z\left(đpcm\right)\)

\(Toru\)