大家帮忙做几道因式分解的题目，谢谢 咯

^2是平方
a^2+(a+1)^2+a^2(a+1)^2
=a^4+2a^3+3a^2+2a+1 (化简部分省去）
=a^4+2a^3+a^2+2a^2+2a+1
=a^2(a^2+2a+1)+2a(a+1)+1
=a^2(a+1)^2+2a(a+1)+1
=[a(a+1)+1]^2
=(a^2+a+1)^2

您看第二行a^4+2a^3+3a^2+2a+1
次数递减，系数1，2，3，2，1
用拆项法3a^2=a^2+2a^2
再用完全平方。

a^4+b^4+(a+b)^4
=2a^4+4a^3b+6a^2b^2+4ab^3+2b^2
=2(a^4+2a^3b+3a^2b^2+2ab^3+b^2)
跟上面比是不是很像？
=2(a^4+2a^3b+a^2b^2+2a^2b^2+2ab^3+b^4)
=2[a^2(a^2+2ab+b^2)+2ab^2(a+b)+b^4]
=2{[a(a+b)]^2+2*a(a+b)*b^2+(b^2)^2}
=2[a(a+b)+b^2]^2
=2(a^2+ab+b^2)^2

下面是个常用的公式：x^3+y^3+z^3-3xyz =(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
竞赛里常会出现，所以一定要记牢。
x^3+y^3+z^3-3xyz
=[(x+y)^3-3xy(x+y)]+z^3-3xyz
=[(x+y)^3+z^3]-3xy(x+y+z)
=[(x+y+z)^3-3(x+y)z(x+y+z)]-3xy(x+y+z)
=(x+y+z)^3-3(xy+yz+zx)(x+y+z)
=(x+y+z)[(x+y+z)^2-3(xy+yz+zx)]
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
要分解他一定要知道x^3+y^3=(x+y)^3-3xy(x+y)这个立方和公式的变形