Question

把(a+b-c)(a-b+c)+(b-a+c)(b-a-c)分解因式

过程!~~~~~~~`

Answer 1

解：(a+b-c)(a-b+c)+(b-a+c)(b-a-c)
=(a+b-c)(a-b+c)-(b-a+c)(a-b+c)
=(a-b+c)[(a+b-c)-(b-a+c)]
=(a-b+c)(2a-2c)
=2(a-b+c)(a-c)

Answer 2

(a+b-c)(a-b+c)+(b-a+c)(b-a-c)
=[a+(b-c)][a-(b-c)]+[(b-a)+c][(b-a)-c]
=a^2-(b-c)^2+(b-a)^2-c^2
=a^2-b^2+2bc-c^2+b^2-2ab+a^2-c^2
=2a^2+2bc-2ab-2c^2
=2(a-b+c)(a-c)

Answer 3

(a+b-c)(a-c+b)+(b-a-c)(b-a-c)
=a^2-ac+ab+ab-bc+b^2-ac+c^2-bc+b^2-ab-bc-ab+a^2+ac-bc+ac+c^2
=2a^2-2ac+2ab-2bc+2b^2+2c^2-2ab-2bc+2ac
=2a^2-4bc+2b^2+2c^2
=2a^2+2(b^2-2bc+c^2)
=2a^2+2(b-c)^2
=2(a^2+(b-c)^2)
(a+b-c)^2+(a+c-b)^2
=
a^2+b^2+c^2+2ab-2ac-2bc+a^2+b^2+c^2+(a+c-b)^2

Answer 4

(a+b-c)(a-b+c)+(b-a+c)(b-a-c)
=(a-b+c)[(a+b-c)-(b-a+c)]
=2(a-b+c)(a-c)