问题标题：
【已知a满足等式a^2-a-1=0,求代数式a^8+7a^-4的值解：由a^2-a-1=0得,a^2=a+1,于是a^4=(a+1)^2=a^2+2a+1=3a+2a^8=(3a+2)^2=9a^2+12a+4=21a+13a^8+7a^-4=21a+13+7/(3a+2)=(63a^2+81a+26+7)/(3a+2)=(63a+63+81a+33)/(3a+2)=(144a+96)/(3a+2)=48请】
问题描述：
已知a满足等式a^2-a-1=0,求代数式a^8+7a^-4的值
解：由a^2-a-1=0得,a^2=a+1,
于是a^4=(a+1)^2=a^2+2a+1=3a+2
a^8=(3a+2)^2=9a^2+12a+4=21a+13
a^8+7a^-4=21a+13+7/(3a+2)
=(63a^2+81a+26+7)/(3a+2)
=(63a+63+81a+33)/(3a+2)
=(144a+96)/(3a+2)=48
请问3a+2和21a+13是怎么来的.
李占斌回答：
a^2+2a+1=(a^2+a)+a+1
又a^2=a+1
∴a^2+2a+1=3a+2
同理
=9a^2+12a+4=9a+12a+4+9=21a+13
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