2x=a+b+c
=>\(x=\dfrac{a+b+c}{2}\)
\(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)\)
\(=\left(x-a\right)\left(x-b+x-c\right)+\left(x-b\right)\left(x-c\right)\)
\(=\left(x-a\right)\left(a+b+c-b-c\right)+\left(x-b\right)\left(x-c\right)\)
\(=\left(\dfrac{a+b+c}{2}-a\right)\cdot a+\left(\dfrac{a+b+c}{2}-b\right)\left(\dfrac{a+b+c}{2}-c\right)\)
\(=\dfrac{-a+b+c}{2}\cdot a+\dfrac{a-b+c}{2}\cdot\dfrac{a+b-c}{2}\)
\(=\dfrac{-a^2+ab+ac}{2}+\dfrac{a^2-\left(b-c\right)^2}{4}\)
\(=\dfrac{-2a^2+2ab+2ac+a^2-b^2+2bc-c^2}{4}\)
\(=\dfrac{-a^2-b^2-c^2+2ab+2ac+2bc}{4}\)
\(ab+bc+ca-x^2\)
\(=\left(ab+bc+ca\right)-\dfrac{\left(a+b+c\right)^2}{4}\)
\(=ab+bc+ca-\dfrac{a^2+b^2+c^2+2ab+2ac+2bc}{4}\)
\(=\dfrac{4ab+4bc+4ac-a^2-b^2-c^2-2ab-2ac-2bc}{4}\)
\(=\dfrac{-a^2-b^2-c^2+2ab+2bc+2ac}{4}\)
Do đó: \(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)=ab+bc+ca-x^2\)
