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Question-159189




Question Number 159189 by mathlove last updated on 14/Nov/21
Answered by Rasheed.Sindhi last updated on 16/Nov/21
Let (A/B)=x  x+(1/x)=1; x^(2019) +x^(−2019) =?  ▶x+(1/x)=1  ⇒x^2 −x+1=0⇒(x+1)(x^2 −x+1)=0  ⇒x^3 +1=0⇒x^3 =−1  ∴ x is cuberoot of −1  ∴ x=−1,−ω,−ω^2   −1 is a root of x+1=0  −ω,−ω^2  are the roots of x^2 −x+1=0  where ω  is cuberoot of unity.  ▶x^(2019) +x^(−2019) =(−ω)^(2019) +(−ω)^(−2019)   −(ω^(2019) +ω^(−2019) )=−({(ω)^3 }^(673) +{(ω)^3 }^(−673) )  =−(1^(673) +1^(673) )=−2  ((A/B))^(2019) +((B/A))^(2019) =−2
$${Let}\:\frac{{A}}{{B}}={x} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1};\:{x}^{\mathrm{2019}} +{x}^{−\mathrm{2019}} =? \\ $$$$\blacktriangleright{x}+\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0}\Rightarrow\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\Rightarrow{x}^{\mathrm{3}} =−\mathrm{1} \\ $$$$\therefore\:{x}\:{is}\:{cuberoot}\:{of}\:−\mathrm{1} \\ $$$$\therefore\:{x}=−\mathrm{1},−\omega,−\omega^{\mathrm{2}} \\ $$$$−\mathrm{1}\:{is}\:{a}\:{root}\:{of}\:{x}+\mathrm{1}=\mathrm{0} \\ $$$$−\omega,−\omega^{\mathrm{2}} \:{are}\:{the}\:{roots}\:{of}\:{x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0} \\ $$$${where}\:\omega\:\:{is}\:{cuberoot}\:{of}\:{unity}. \\ $$$$\blacktriangleright{x}^{\mathrm{2019}} +{x}^{−\mathrm{2019}} =\left(−\omega\right)^{\mathrm{2019}} +\left(−\omega\right)^{−\mathrm{2019}} \\ $$$$−\left(\omega^{\mathrm{2019}} +\omega^{−\mathrm{2019}} \right)=−\left(\left\{\left(\omega\right)^{\mathrm{3}} \right\}^{\mathrm{673}} +\left\{\left(\omega\right)^{\mathrm{3}} \right\}^{−\mathrm{673}} \right) \\ $$$$=−\left(\mathrm{1}^{\mathrm{673}} +\mathrm{1}^{\mathrm{673}} \right)=−\mathrm{2} \\ $$$$\left(\frac{{A}}{{B}}\right)^{\mathrm{2019}} +\left(\frac{{B}}{{A}}\right)^{\mathrm{2019}} =−\mathrm{2} \\ $$

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