c1-c3,c2-c3
(a+b+c)(a+b-c) 0 c^2
0 (a+b+c)(b+c-a) a^2
(a+b+c)(b-a-c) (a+b+c)(b-a-c) (c+a)^2
1,2列分别提出a+b+c
a+b-c 0 c^2
0 b+c-a a^2
b-a-c b-a-c (c+a)^2
r3-r1-r2
a+b-c 0 c^2
0 b+c-a a^2
-2a -2c 2ac
c3+cc1
a+b-c 0 ac+bc
0 b+c-a a^2
-2a -2c 0
= 2(a+b+c)^2[ac(a+b)(b+c-a)+ca^2(a+b-c)]
= 2ac(a+b+c)^2(ab+ac-a^2+b^2+bc-ab + a^2+ab-ac)
= 2ac(a+b+c)^2(b^2+bc+ab)
= 2abc(a+b+c)^3