Question

已知a^2+3a+1=0,求3a^2+(a^2+5)(a^2-1)-a(5a+6)的值

Answer 1

a^2+3a+1=0

a^2+3a=-1

3a^3+(a^2+5)(a^2-1)-a(5a+6)

=3a^3+a^4+4a^2-5-5a^2-6a

=a^4+3a^3+4a^2-5-5a^2-6a

=a^2(a^2+3a)+4a^2-5-5a^2-6a

=-a^2+4a^2-5-5a^2-6a

=-2a^2-6a-5

=-2(a^2+3a)-5

=-2*(-1)-5

=2-5

= -3