a(x-b)\left(x-c \right)+b\left(x-c \right)\left(x-a \right)+c\left(x-a \right)\left(x-b \right)=0
\Leftrightarrow \left a(x^{2}-bx-ax+bc \right)+ b\left(x^{2}-cx-ax+ca \right)+c \left(x^{2}-ax-bx+ab \right)\Leftrightarrow ax^{2}-abx-cax+abc+bx^{2}-bcx-abx+abc+cx^{2}-cax-bcx+abc=0\Leftrightarrow \left(a+b+c \right)x^{2}-2\left(ab+ac+bc \right)x+3abc
\Rightarrow \Delta ' \left(ab+ac+bc \right)^{2}-3abc\left(a+b+c \right)=a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+2a^{2}bc+2ab^{2}c+2abc^{2}-3a^{2}bc-3ab^{2}c-3abc^{2}=a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}-a^{2}bc-ab^{2}c-abc^{2}. nhân hai vế với 2 ta có:2\Delta =2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}-2a^{2}bc-2ab^{2}c-2abc^{2}=\left(a^{2}b^{2}-2a^{2}bc+a^{2}c^{2} \right)+\left(a^{2}b^{2}-ab^{2}c+b^{2}c^{2} \right)+\left(b^{2}c^{2}-2abc^{2}+a^{2}c^{2} \right)=\left(ab-ac \right)^{2}+\left(ab-bc \right)^{2}+\left(bc-ac \right)^{2}
\Rightarrow \Delta \geq 0 \Rightarrow phương trình luôn có nghiệm