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