Ta có: \(S=a+b+c\left(1\right)\)
Thay \(\left(1\right)\)vào ta được:
\(\left(S-2b\right).\left(S-2c\right)=\left(a+b+c-2b\right).\)\(\left(a+b+c-2c\right)\)
\(=\left(a-b+c\right).\left(a+b-c\right)\)
\(=a^2+ab-ac-ba-b^2+bc+ca+cb-c^2\)
\(=a^2-b^2-c^2+2.bc\left(2\right)\)
Tương tự, ta được:
\(\left(S-2c\right).\left(S-2a\right)=b^2-c^2-a^2+2.ca\left(3\right)\)
\(\left(S-2a\right).\left(S-2b\right)=c^2-a^2-b^2+2.ab\left(4\right)\)
Từ \(\left(2\right);\left(3\right);\left(4\right)\Rightarrow\)Tổng bằng:
\(a^2-b^2-c^2+2bc+b^2-c^2-a^2+2ca+c^2-a^2\)\(-b^2+2ab\)
\(=2ab+2bc+2ca-a^2-b^2-c^2\)
Vậy tổng trên \(=2ab+2bc+2ca-a^2-b^2-c^2.\)
Thay S=a+b+c vào biểu thức ta được:
(a+b+c-2b)(a+b+c-2c)+(a+b+c-2c)(a+b+c-2a)+(a+b+c-2a)(a+b+c-2b)
=(a-b+c)(a+b-c)+(b-c+a)(b+c-a)+(c-a+b)(c+a-b)
=a2-(b-c)2+b2-(c-a)2+c2-(a-b)2
=a2-b2+2bc-c2+b2-c2+2ac-a2+c2-a2+2ab-b2
=-a2-b2-c2+2ab+2bc+2ca
