条件给出:x/(x^2-mx+1)=1
两边的倒数也相等
(x^2-mx+1)/x=1
x+1/x-m=1
x+1/x=m+1
两边同时三次方,即
(x+1/x)^3=(1+m)^3
x^3+1/x^3+3(x+1/x)=m^3+3m^2+3m+1
x^3+1/x^3=m^3+3m^2+3m+1-3(x+1/x)
=m^3+3m^2+3m+1-3m-3
=m^3+3m^2-2
另外,要求的式子:
x^3/(x^6-(mx)^3+1)的倒数为
x^3-m^3+1/x^3 = x^3+1/x^3-m^3
= m^3+3m^2-2-m^3
= 3m^2-2
最后得:x^3/(x^6-(mx)^3+1)=1/(3m^2-2)