Question

已知a+b=1,a^3+3a^2+3a+3b-3b^2+b^3=37求(a+1)^4+(b-1)^4

Answer 1

∵a+b=1
∴b=1-a
∵a^3+3a^2+3a+3b-3b^2+b^3=37

∴a^3+3a^2+3a+3b-3b^2+b^3
=（a^3+b^3)+3(a^2-b^2)+3(a+b)
=(a+b)(a^2-ab+b^2)+3(a+b)(a-b)+3
=a^2-ab+b^2+3a-3b+3
=(a^2+2ab+b^2)-3ab+3a-3b+3
=（a+b)^2-3a(1-a)+3a-3(1-a)+3
=1+3a^2-3a+3a+3a
=3a^2+3a+1
=37
∴3a^2+3a-36=0
∴a^2+a-12=0
解得a=-4,b=5
或a=3,b=-2

若a=-4,b=5
(a+1)^4+(b-1)^4
=3^4+4^4=81+256=337

若a=3,b=-2
（a+1)^4+(b-1)^4
=4^4+(-3)^4=337
总有
(a+1)^4+(b-1)^4=337