设x/(x^2-mx+1)=1,其中m为常数,求x ^3/(x^6-m^3x^3+1)的值详细过程

题目

设x/(x^2-mx+1)=1,其中m为常数,求x ^3/(x^6-m^3x^3+1)的值详细过程

答案

设x/(x^2-mx+1)=1,其中m为常数,求x ^3/(x^6-m^3x^3+1)的值详细过程
x/(x^2-mx+1)=1,即(x^2-mx+1)/x=1,即 x + 1/x = 1+m
设x^3/(x^6-m^3x^3+1)的倒数为y
y = (x^6-m^3x^3+1)/x^3
= x^3 + 1/x^3 - m^3
= (x + 1/x)(x^2 + 1/x^2 - 1) - m^3
= (x + 1/x)[(x + 1/x)^2 - 3] - m^3
= (1+m)[(1+m)^2-3] - m^3
= 3m^2 - 2
所以 x^3/(x^6-m^3x^3+1) = 1/y = 1/(3m^2 - 2)