Ta có : \(a\left(b-c\right)\left(b+c-a\right)^2+c\left(a-b\right)\left(a+b-c\right)^2-b\left(a-c\right)\left(a+c-b\right)^2=0\left(1\right)\)
Đặt : \(\left[{}\begin{matrix}a+b-c=x\\b+c-a=y\\a+c-b=z\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=\dfrac{x+z}{2}\\b=\dfrac{x+y}{2}\\c=\dfrac{y+z}{2}\end{matrix}\right.\)
Khi đó ta có :
\(VT_{\left(1\right)}=\dfrac{x+z}{2}\left(\dfrac{x+y}{2}-\dfrac{y+z}{2}\right).y^2+\dfrac{y+z}{2}\left(\dfrac{x+z}{2}+\dfrac{x+y}{2}\right).x^2-\dfrac{1}{4}\left(x+y\right)\left(x-y\right).z^2\)
\(=\dfrac{x+z}{2}.\dfrac{x-z}{2}.y^2+\dfrac{y+z}{2}.\dfrac{z-y}{2}.x^2+\dfrac{1}{4}\left(x^2-y^2\right)z^2\)
\(=\dfrac{1}{4}\left(x^2-z^2\right).z^2-\dfrac{1}{4}\left(x^2-y^2\right).z^2=0\left(đpcm\right)\)
