If #x+1/x=1#, what is the value of #x^2001+1/x^2001# ?

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Answer 1 · 2018-06-01T13:38:19

Given
#x+1/x=1#,

#=>x^2+1=x#

#=>x^2-x+1=0#

We get

#x^3=x^3+1-1=(x+1)(x^2-x+1)-1=(x+1)xx0-1=-1#

Now

#x^2001+1/x^2001#

#=(x^3)^667+1/(x^3)^667#

#=(-1)^667+1/(-1)^667=-1-1=-2#