已知a-1/a=√3,求(a^10+a^6+a^4+1)/(a^10+a^8+a^2+1)

Question

已知a-1/a=√3,求(a^10+a^6+a^4+1)/(a^10+a^8+a^2+1)

Answer 1

a-1/a=√3
两边平方：a²+1/a²-2*a*1/a=3
a²+1/a²-2=3
a²+1/a²=5
两边平方:
a^4+1/a^4+2=25
a^4+1/a^4=23
(a^10+a^6+a^4+1)/(a^10+a^8+a^2+1)
=(a^6+1)(a^4+1)/[(a^2+1)(a^8+1)]
=(a^2+1)(a^4-a^2+1)(a^4+1)/[(a^2+1)(a^8+1)],
=(a^4-a^2+1)(a^4+1)/(a^8+1)
=(a^4-a^2+1)(a^2+1/a^2)/(a^6+1/a^2),【分子分母同除以a^2】
=(a^4-a^2+1)(5)/(a^6+1/a^2)
=5(a^2-1+1/a^2)/(a^4+1/a^4),【分子分母同除以a^2】
=5(5-1)/23
=20/23