a^-3a+1=0,求2a^5-5a^4+2a^3-8a^2+3a得多少,请写出计算过程,

a^2-3a+1=0
所以a^2=3a-1
所以a^3=a^2*a=(3a-1)*a=3a^2-a=3(3a-1)-a=8a-3
a^4=(a^2)^2=9a^2-6a+1=9(3a-1)-6a+1=21a-8
a^5=a^4*a=(21a-8)*a=21a^2-8a=21(3a-1)-8a=55a-21
所以
2a^5-5a^4+2a^3-8a^2+3a
=2(55a-21)-5(21a-8)+2(8a-3)-8(3a-1)+3a
=110a-42-105a+40+16a-6-24a+8+3a
=0

2a^5 - 5a^4 + 2a^3 - 8a^2 + 3a
= 2a^3 (a^2 - 3a + 1) + a^4 - 8a^2 + 3a
= a^4 - 8a^2 + 3a
= a^2( a^2 -3a + 1) + 3a^3 - 9a^2 + 3a
= 3a^3 - 9a^2 + 3a
= 3a(a^2 -3a + 1)
=0