a^3+b^3+3a^2+3b^2+3a+3b+2=

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a^3+b^3+3a^2+3b^2+3a+3b+2=a^3+b^3+3a^2+3b^2+3a+3b+2=a^3+b^3+3a^2+3b^2+3a+3b+2=原式=(a³+3a²+3

a^3+b^3+3a^2+3b^2+3a+3b+2=
a^3+b^3+3a^2+3b^2+3a+3b+2=

a^3+b^3+3a^2+3b^2+3a+3b+2=
原式=(a³+3a²+3a+1)+(b³+3b²+3b+1)
=(a+1)³+(b+1)³
=[(a+1)+(b+1)][(a+1)²-(a+1)(b+1)+(b+1)²]
=(a+b+2)(a²-ab+b²+a+b+1)

a^3+b^3+3a^2+3b^2+3a+3b+2
=a^3+3a^2+3a+1+b^3+3b^2+3b+1
=(a+1)^3+(b+1)^3
=[(a+1)+(b+1)][(a+1)^2-(a+1)(b+1)+(b+1)^2]
=(a+b+2)[(a+1)^2-(a+1)(b+1)+(b+1)^2]

a^3+b^3+3a^2+3b^2+3a+3b+1=(a+b)^3二项式定理直接可以看出
因为x^3+y^3=(x+y)(x^2+y^2-xy)
在这里x=a+b y=1带入即可
(a+b)^3+1=(a+b+1)(a^2+2ab+b^2-a-b+1)