设x-1/x=3,求x^10+x^8+x^2+1/x^10+x^6+x^4+1的值.10次方,8次方,不好打出来-后面那个是分数（“/”）!

Question

设x-1/x=3,求x^10+x^8+x^2+1/x^10+x^6+x^4+1的值
.10次方,8次方,不好打出来- 后面那个是分数（“/”）!

Answer 1

(x^10+x^8+x^2+1)/(x^10+x^6+x^4+1)
=(x^2+1)(x^8+1)/(x^4+1)(x^6+1)
=[(x^2+1)/x]*[(x^8+1)/x^4]/[(x^4+1)/x^2]*[(x^6+1)/x^3]
=[(x+1/x)(x^4+1/x^4)]/[(x^2+1/x^2)(x^3+1/x^3)]
=(x+1/x)[(x^2+1/x^2)^2-2]/[(x-1/x)^2+2][(x+1/x)(x^2-1+1/x^2)]
=(x+1/x)[[(x-1/x)^2+2]^2-2}/[(x-1/x)^2+2](x+1/x)[(x-1/x)^2+1]
=(x+1/x)[(3^2+2)^2-2]/(3^2+2)(x+1/x)(3^2+1)
=119(x+1/x)/110(x+1/x)
=119/110