已知：X^3+Y^(-3)=2,求X+X^(-1)的值

设X+X^(-1)=t
X^3+X^(-3)
=[X+X^(-1)][X^2+X^(-2)-1]
=t[(X+X^(-1))^2-2-1]
=t(t^2-3)=2
即t^3-3t-2=0
t(t^2-1)-2(t+1)=0
(t+1)(t^2+t-2)=0
(t+1)^2(t-2)=0
所以t=-1,t=2

是X^3+X^(-3)=2吧？
设X+X^(-1)=t
X^3+X^(-3)
=[X+X^(-1)][X^2+X^(-2)-1]
=t[(X+X^(-1))^2-2-1]
=t(t^2-3)=2
即t^3-3t-2=0
方程自己解吧

X^3+X^(-3)

=(X+1/X)(X^2-1+1/X^2)

=(X+1/X)[(X+1/X)^2-3]

=(X+1/X)^3-3(X+1/X)=2

即(X+1/X)^3-3(X+1/X)-2=0

所以，
（X+1/X)3+(X+1/X)^2-(X+1/X)^2-3(X+1/X)-2=0

(X+1/X)^2[(X+1/X)+1]-[(X+1/X)^2+3(X+1/X)+2]=0

(X+1/X)^2[(X+1/X)+1]-[(X+1/X)+1][(X+1/X)+2]=0

(X+1/X+1)[(X+1/X)^2-(X+1/X)-2]=0

(X+1/X+1)(X+1/X-2)(X+1/X+1)=0

所以，X+1/X=-1或X+1/X=2