Question

مطلوب است محاسبه $ ( \frac{1}{ x^{2}+x+1 } + \frac{1}{ x^{2}-x+1 } )^5 $در صورتی که $1= x^{5} $و$x \neq 1$ از اتحاد $( x^{4} + x^{3}+ x^{2}+x+1) (x-1)= x^{5}-1 $ استفاده کردم.

Answer 1

$x \neq 1 \wedge x^5=1 \wedge (x^4+x^3+x^2+x+1)(x-1)=x^5-1 \Rightarrow x^4+x^3+x^2+x+1=0$

$ \Rightarrow x^4+x^2+1=-(x^3+x)=-x(x^2+1)$

$ \Rightarrow ( \frac{1}{x^2+x+1} + \frac{1}{x^2-x+1} )^5=( \frac{x^2-x+1+x^2+x+1}{(x^2+x+1)(x^2-x+1)} )^5=( \frac{x^2-x+1+x^2+x+1}{(x^2+1)^2-x^2} )^5=( \frac{2(x^2+1)}{x^4+x^2+1} )^5$

$=( \frac{2(x^2+1)}{-x(x^2+1)} )^5=(\frac{2}{-x})^5= \frac{-2^5}{x^5} $

$= \frac{-2^5}{1}=-2^5=-32$

$ \Box $

Answer 2

$$ x^{5} -1=0 \Longrightarrow (x-1)( x^{4} + x^{3} + x^{2} +x+1)=0,x \ne1 \Longrightarrow x^{4} + x^{3} + x^{2}+ x+1=0 \Longrightarrow x^{4} + x^{2} +1=-x( x^{2} +1) \Longrightarrow ( \frac{1}{ x^{2} +x+1} + \frac{1}{ x^{2} -x+1} )^{5} = ( \frac{2( x^{2} +1)}{ x^{4} + x^{2} +1} )^{5} = ( \frac{2( x^{2} +1)}{-x( x^{2} +1)} )^{5}= ( \frac{-2}{x} )^{5}= \frac{-32}{ x^{5} }=-32 $$