已知a²+3a+1=0,求代数式3a³+（a²+5）（a²-1）-5（a+1)(a-1)-6a
问题描述：
已知a²+3a+1=0,求代数式3a³+（a²+5）（a²-1）-5（a+1)(a-1)-6a
答
∵a²+3a+1=0,∴a²+3a=-1,
∴3a³+(a²+5)(a²-1)-5(a+1)(a-1)-6a
=3a³+(a²+5)(a²-1)-5(a²-1)-6a
=3a³+a²(a²-1)-6a
=a^4+3a³-a²-6a
=a²(a²+3a)-a²-6a
=-2a²-6a
=-2(a²+3a)
=2.