Question Number 187971 by mnjuly1970 last updated on 24/Feb/23
$$ \\ $$$$\:\:\mathrm{If}\:\:\:,\:\:{x}^{\:\mathrm{5}} \:=\:\mathrm{1}\:\:\:\wedge\:\:{x}\neq\mathrm{1} \\ $$$$ \\ $$$$\:\:\:\:\left(\:\frac{\:\mathrm{1}}{{x}^{\:\mathrm{2}} \:−{x}\:+\mathrm{1}}\:+\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} \:+\:{x}\:+\mathrm{1}}\:\right)^{\:\mathrm{10}} =\:? \\ $$$$ \\ $$
Answered by BaliramKumar last updated on 24/Feb/23
$$\mathrm{2}^{\mathrm{10}} \:=\:\mathrm{1024} \\ $$
Answered by Rasheed.Sindhi last updated on 25/Feb/23
![If x^( 5) = 1 ∧ x≠1 ( (( 1)/(x^( 2) −x +1)) + (1/(x^( 2) + x +1)) )^( 10) = ? x^5 −1=0 (x−1)(x^4 +x^3 +x^2 +x+1)=0 x^4 +x^3 +x^2 +x+1=0 [∵ x≠1] x^2 +(1/x^2 )+x+(1/x)+1=0 [x≠0] x^2 +(1/x^2 )+1=−(x+(1/x)).....A ▶( (( 1)/(x^( 2) −x +1)) + (1/(x^( 2) + x +1)) )^( 10) ( (( 1)/(x^ +(1/x)−1)) + (1/(x^ + (1/x) +1)) )^( 10) (((2(x+(1/x)))/((x+(1/x))^2 −1)))^(10) (((2(x+(1/x)))/(x^2 +(1/x^2 )+1)))^(10) From A (((2(x+(1/x)))/(−(x+(1/x)))))^(10) [∵ x^2 +(1/x^2 )+1=−(x+(1/x))...from A] =(−2)^(10) =2^(10) =1024](https://www.tinkutara.com/question/Q188050.png)
$$\:\:\mathrm{If}\:\:\:\:{x}^{\:\mathrm{5}} \:=\:\mathrm{1}\:\:\:\wedge\:\:{x}\neq\mathrm{1} \\ $$$$\left(\:\frac{\:\mathrm{1}}{{x}^{\:\mathrm{2}} \:−{x}\:+\mathrm{1}}\:+\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} \:+\:{x}\:+\mathrm{1}}\:\right)^{\:\mathrm{10}} =\:? \\ $$$$ \\ $$$${x}^{\mathrm{5}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}\:\:\left[\because\:{x}\neq\mathrm{1}\right] \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}=\mathrm{0}\:\:\:\:\left[{x}\neq\mathrm{0}\right] \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}=−\left({x}+\frac{\mathrm{1}}{{x}}\right)…..{A} \\ $$$$ \\ $$$$\blacktriangleright\left(\:\frac{\:\mathrm{1}}{{x}^{\:\mathrm{2}} \:−{x}\:+\mathrm{1}}\:+\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} \:+\:{x}\:+\mathrm{1}}\:\right)^{\:\mathrm{10}} \\ $$$$\left(\:\frac{\:\mathrm{1}}{{x}^{\:} \:\:+\frac{\mathrm{1}}{{x}}−\mathrm{1}}\:+\:\frac{\mathrm{1}}{{x}^{\:} \:+\:\frac{\mathrm{1}}{{x}}\:+\mathrm{1}}\:\right)^{\:\mathrm{10}} \\ $$$$\left(\frac{\mathrm{2}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{10}} \\ $$$$\left(\frac{\mathrm{2}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}}\right)^{\mathrm{10}} \:\:{From}\:\:{A} \\ $$$$\left(\frac{\mathrm{2}\cancel{\left({x}+\frac{\mathrm{1}}{{x}}\right)}}{−\cancel{\left({x}+\frac{\mathrm{1}}{{x}}\right)}}\right)^{\mathrm{10}} \left[\because\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}=−\left({x}+\frac{\mathrm{1}}{{x}}\right)…{from}\:{A}\right] \\ $$$$=\left(−\mathrm{2}\right)^{\mathrm{10}} =\mathrm{2}^{\mathrm{10}} =\mathrm{1024} \\ $$