若a²+a-3=0,求a⁴+2a³-a-1

若a²+a-3=0,求a⁴+2a³-a-1
若a²+a-3=0,求a⁴+2a³-a-1

若a²+a-3=0,求a⁴+2a³-a-1
a²+a=3
原式=a⁴+a³+a³-a-1
=a²(a²+a)+a³-a-1
=3a²+a³-a-1
=a³+a²+2a²-a-1
=a(a²+a)+2a²-a-1
=3a+2a²-a-1
=2(a²+a)-1
=2×3-1
=5

a²+a-3=0
则
a²+a=3
则
原式
=a⁴+2a³-a-1
=a²(a²+1)+a³-a-1
=3a²+a³-a-1
=a(a²+a)+2a²-a-1
=3a+2a²-a-1
=2a²+2a-1
=2(a²+a)-1
=2*3-1
=5

∵a²+a-3=0
a²+a=3
∴a⁴+2a³-a-1
=a²(a²+a)+a³-a-1
=a³+3a³-a-1
=a(a²+a)+2a²-a-1
=2a²+3a-a-1
=2(a²+a)-1
=6-1
=5