Question

一道因式分解

2a2b2＋2a2c2＋2b2c2－a4－b4－c4

Answer 1

2a2b2＋2a2c2＋2b2c2－a4－b4－c4=-(a4+b4+c4-2a2b2-2b2c2-2a2c2)
=-(a4+b4+c4-2a2b2+2b2c2-2a2c2-4b2c2)
=-[(a2-b2-c2)2-4b2c2]
=-(a2-b2-c2+2bc)(a2-b2-c2-2bc)
=-[a2-(b-c)2][a2-(b+c)2]
=-(a-b+c)(a+b-c)(a-b-c)(a+b+c)

Answer 2

2a2b2＋2a2c2＋2b2c2－a4－b4－c4=2a2b2＋2a2c2＋2b2c2－2a4－2b4－2c4+a4+b4+c4=-(a4+b4-2a2b2+b4+c4-2b2c2+a4+c4-2a2c2-c4-b4-a4)=-(a2-b2)2-(b2-c2)2-(a2-c2)2+c4+b4+a4 = -{(a2-b2)2 - c4+ (a2-c2)2 -b4 -(b2-c2)2 -a4}=-{(a2-c2-b2) (a2-b2+c2) +(a2-c2-b2) (a2-c2+b2) +(b2-c2-a2) (b2-c2+a2)}=-(a2-c2-b2)(3a2+b2-c2)