一道初二数学题不会做...

已知x^2/x^4+1=1/3 , 求x^4/x^8+x^4+1的值
x^2/(x^4+1)=1/3
取倒数:(x^4+1)/x^2=3
x^2+1/x^2=3

那么:(x^8+x^4+1)/x^4=x^4+1+1/x^4
=(x^2+1/x^2)^2-1
=3^2-1
=8
所以,x^4/(x^8+x^4+1)=1/8

x^2/(x^4+1)=1/3
所以(x^4+1)/x^2=x^2+1/x^2=3
x^4/(x^8+x^4+1)
=1/[[x^8+x^4+1)/x^4]
=1/(x^4+1+1/x^4)
=1/[(x^2)^2+2*x^2*(1/x^2)-2+1+(1/x^2)^2]
=1/[(x^2+1/x^2)^2-1]
=1/(3^2-1)
=1/8

x^2/x^4+1=1/3
1/x^2=-2/3
1/x^4=4/9
x^4=9/4

x^4/x^8+x^4+1
=1/x^4+x^4
=4/9+9/4
=97/36

答：
3x2=x4+1平方
9x4=x8+2x4+1
即8x4=x8+x4+1
x4/(x8+x4+1)=1/8

x^2=1 (x^4+1)=3 x^4=2(x^2)=2 (x^8+x^4+1)=(x^12+1) x^4=2
(x^12+1)=7 求x^4/(x^8+x^4+1)=2/7