If #x+1/x = -1# then what is #x^2018+1/x^2018# ?

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Answer 1 · 2018-02-15T23:21:22

#x^2018+1/x^2018 = -1#

Given:

#x+1/x = -1#

Multiply both sides by #x# to get:

#x^2+1 = -x#

Add #x# to both sides to get:

#x^2+x+1 = 0#

Multiply by #(x-1)# to get:

#x^3-1 = 0#

So the roots of #x^2+x+1=0# are both cube roots of #1#

Note that #2019 = 3 * 673#

So:

#x^2018+1/x^2018 = x^2019/x + x/x^2019 = (x^3)^673/x+x/((x^3)^673) = 1/x+x = -1#

Answer 2 · 2018-02-15T23:22:15

# x^2018 + x^(-2018) = -1 #

For this problem I assume you understand De Moivres Theorem of complex numbers...

if #z = costheta + i sin theta #

#=> (costheta + isintheta)^n -= cosntheta + isinntheta #

#=> z^n + z^(-n) = 2cosntheta #

So

#x+x^(-1) = -1 #

Letting #x = costheta + isintheta #

#=> 2costheta = -1 #

#=> x^2018 + x^(-2018) = 2cos2018theta #

Now we must find #2cos2018theta #...

#2costheta = -1 => theta = (2pi)/3 #

#therefore 2cos2018theta = 2cos(2018*(2pi)/3) = -1 #

#therefore x^2018 + x^(-2018) = -1 #