2道超难的因式分解

1题

解：设y-z=a, z-x=b, x-y=c,则
a+b+c=0，
y+z-2x=b-c
x+z-2y=c-a
x+y-2z=a-b
于是
原式=（y+z-2x)^2+(z+x-2y)^2+(x+y-2z)^2
=(b-c)^2+(c-a)^2+(a-b)^2-3a^2-3b^2-3c^2
=b^2-2ac+c^2+c^2-2ac+a^2+a^2-2ab+b^2-3a^2-3b^2-3c^2
=-a^2 -b^2-c^2-2ac-2ab-2bc
=-(a^2 +b^2+c^2+2ac+2ab+2bc)
=-[(a+b)^2+2c(a+b)+c^2]
=-(a+b+c)^2
=0

2题
解：
a^3+b^3 +c^3-a^2b-a^2c-ab^2-b^2c-ac^2-bc^2 +2abc
= a^3+b^3 +c^3-a (b^2+c^2)-b (a^2+c^2)-c (a^2+b^2)+2abc
= a^3-a (b^2+c^2)+b^3 -b (a^2+c^2)+c^3-c (a^2+b^2)+2abc
=a(a^2-b^2-c^2)+b(b^2-a^2-c^2)+c(c^2-a^2-b^2) +2abc+2abc-2abc
=a(a^2-b^2+2bc-c^2) +b(b^2-a^2-c^2+2ac)+c(c^2-a^2-b^2-2ab)
=a[a^2-(b-c)^2]+b[b^2-(a-c)^2]+c[(c^2-(a+b)^2]
＝a(a-b+c)(a+b-c)+b(b-a+c)(b+a-c)+c(c-a-b)(c+a+b)
= a(a-b+c)(a+b-c)+b(b-a+c)(b+a-c)-c(a+b-c)(c+a+b)
=(a+b-c)[a(a-b+c)+b(b-a+c)-c(a+b+c)]
=(a+b-c)[a^2-ab+ac+b^2-ab+bc-ac-bc-c^2]
=(a+b-c)[a^2-2ab +b^2 -c^2]
=(a+b-c)[