Question

一道数学计算题

已知a^2-a-1=0,求a^18+323a^-6的值....

Answer 1

a^18+(323/a^6)
=(a^2)^9+(323/(a^2)^3)
=(a+1)^9+(323/(a+1)^3)
=A^3+(323/A)
=(A^4+323)/A
其中，A=(a+1)^3

根据已知条件：
a^2-a-1=0
将其化为：(a+1)^2-3(a+1)+1=0
即：(a+1)^2=3(a+1)-1
(a+1)^3=3(a+1)^2-(a+1)=8(a+1)-3
即：A=8(a+1)-3
A^2=64(a+1)^2-48(a+1)+9
=192(a+1)-64-48(a+1)+9
=144(a+1)-55
A^4=20736(a+1)^2-15840(a+1)+3025
=62208(a+1)-20736-15840(a+1)+3025
=46368(a+1)-17711

原式
=(A^4+323)/A
=(46368(a+1)-17711+323)/(8(a+1)-3)
=(46368a+28980)/(8a+5)
=5796*(8a+5)/(8a+5)
=5796

答：已知a^2-a-1=0，则a^18+(323/a^6)=5796