已知x^4-3x^2+1=0,求x^8+7/x^4的值

已知x^4-3x^2+1=0,求x^8+7/x^4的值
2）已知x^2-4x+1=0,求x^3+4/x^2-1/x的值

一
1）x^8=(x^4)^2=(3x^2-1)^2
2）x^4-3x^2+1=0 等式两边同除以x^2,则x^2-3+1/x^2=0,故（x^2+1/x^2）^2=9即x^4+1/x^4=7
故x^8+7/x^4=9x^4-6x^2+1+7/x^4
=2x^4-6x^2+1+(1-1)+7x^4+7/x^4
=0-1+7*7
=48
二
1）同除以x得：x+1/x=4
同时平方：x^2+1/x^2=14
2）x^3+4/x^2-1/x=x^3+(-4x^2+x+4x^2-x)+4/x^2-1/x
=x*(x^2-4x+1)+4x^2-x+4/x^2-1/x
=0+4(x^2+1/x^2)-(x+1/x)
=4*14-4
=52