Question

若X/(x2-mx+1)=1,求x3/(x6-m3x3+1)的值

Answer 1

X/(x2-mx+1)=1 左边分子分母同除以X,得 1/[X-m+(1/X)]=1 故X+1/X=m+1 x3/(x6-m3x3+1) =1/[x^3-m^3+(1/x^3)] =1/[(x+1/x)(x^2+1/x^2-1)-m^3] 因x^2+1/x^2=(x+1/x)^2-2 故原式=1/{（m+1）[(m+1)^2-2-1)]-m^3} =1/(m+1)^3-3(m+1)-m^3 =1/m^3+3m^2+3m+1-3m-3-m^3 =1/3m^2-2