If #x+1/x=-1# then What is the value of #x^247+1/x^187=#?

As already obtained in the first answer,

#x=-1/2+-i sqrt 3/2=(r (cos theta+i sin theta))=re^(i theta), respectively#

r = 1.

For the first root, cosine is negative and sine is positive. #theta# is in

the second quadrant and is #(2pi)/3#.

For the second, both are negative. #theta# is in the 3rd quadrant.

So, it is #(4pi)/3#

Now, #x = e^((2pi)/3i) and e^((4pi)/3i)#

Accordingly,

#x^247+1/x^247#

#= e^(247(2pi)/3i)+e^(-247(2pi)/3i)# and

#e^(247(4pi)/3i)+e^(-247(4pi)/3i)#

#=2 cos (247(2pi)/3i) and 2 cos (247(4pi)/3i)#

#=2 cos (164pi+(2pi)/3) and 2 cos (329pi+pi/3)#

=2 cos (even #pi + (2pi)/3#) and 2 cos (odd #pi+pi/3#)

#=2 cos ((2pi)/3) and -2 cos (pi/3)#

#=2(-1/2) and (-2)(1/2)#

Both are the same -1...

In the equation #x+1/x=-1# occurs #abs x=1# because making #x = 1+delta# and substituting there are not real solutions for #delta# in the equation #1+delta + 1/(1+delta)=-1#: then we can make #x = e^{i theta}#.

Substituting we have

#e^{i theta} + e^{- i theta} = -1#

but

#(e^{i theta} + e^{- i theta} )/2 = cos(theta)#

so the equation reduces to

#2 cos(theta)=-1#. Solving for #theta# we have:

#theta = pm 2/3 pi+2k pi# with #k = 0, 1,2,...#

Now #x^{247} = e^{pm i times 247 times2/3 pi} = e^{pm i2/3 pi}#

and

#x^{187} = e^{pm i times 187 times 2/3 pi} = e^{pm i2/3 pi}#

Putting all together

#x^{247} +1/(x^{187}) = 2 cos(pm 2/3 pi) = 2(-1/2)=-1#

For lower class student I do it as follows

Given

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

So

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

Now

#x^247+1/x^187=(x^3)^82 *x+1/((x^3)^62*x)=(1^82*x)+1/(1^62*x)=x+1/x=-1#

Will this do?