Question

一道数学问题求解

已知x^2-3x+1=0,求x^3+1/x^3值

Answer 1

x^2-3x+1=0
x^2+1=3x
x+1/x=3
x^3+1/x^3
=(x+1/x)(x^2-1+1/x^2)
=3(x^2+1/x^2-1)
=3[(x+1/x)^2-2-1]
=3*(9-3)
=18

Answer 2

解：
x=0显然不是方程的解
两边除以x
故x-3+1/x=0
x+1/x=3
两边同时平方
x^2+1/x^2+2=9
所以x^3+1/x^3=（x+1/x）*（x^2+1/x^2-1）=18

Answer 3

x^2-3x+1=0
x+1/x=3
（x+1/x）^2=9
x^2+1/x^2=7
问题x^3+1/x^3=（x+1/x）（x^2+1/x^2-1）=3*（7-1）=18

Answer 4

x^2-3x+1=0的两根为x1,x2
x1*x2=1
x1^3=1/x2^3
所以x1^3+1/x1^3=x2^3+1/x2^3=x1^3+x2^3=（x1+x2）(x1^2-x1*x2+x2^2)
x1+x2=3 x1*x2=1 x1^2+x2^2=3^2-2*1=7
原始=3*（7-1）=18