已知x/(x^2+x+1)=a,求分式x^2/(x^4+x^2+1)的值

x/(x^2+x+1)=a,1/[x+1+(1/x)]=a,x+(1/x)+1=1/a,x+(1/x)=(1-a)/a,x^2/(x^4+x^2+1)=1/[x^2+1+(1/x^2)]=1/{[x+(1/x)]^2-2*x*(1/x)+1}=1/{[(1-a)/a]^2-1}=1/[(1-a)^2/a^2-a^2/a^2]=1/[(1-2a)/a^2]=a^2/(1-2a).