已知a+b=4,ab=1,求(a的平方+4b+1)(b的平方+4a+1)的值

Question

已知a+b=4,ab=1,求(a的平方+4b+1)(b的平方+4a+1)的值

Answer 1

解原式
a^2b^2+4a^3+a^2+4b^3+16ab+4b+b^2+4a+1
=1^2+4a^3+4b^3+a^2+16ab+b^2+4b+4a+1
=1+4(a^3+b^3)+a^2+b^2+16ab+4(a+b)+1
=4(a^3+b^3)+a^2+b^2+16ab+4(a+b)+2
=4(a+b)(a^2-ab+b^2)+(a+b)^2-2ab+16ab+4(a+b)+2
=4*4(a^2+b^2-ab)+(4)^2-2*1+16*1+4*4+2
=16[(a+b)^2-3ab]+16-2+16+16+2
=16[(4)^2-3]+16-2+16+16+2
=16*13+16*3
=16*16
=256