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Question Number 185706 by Shrinava last updated on 26/Jan/23

if x^2 + x + 1 = 0 find: x^(2011) + (1/x^(2011) ) = ?

$$\mathrm{if}\:\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{find}:\:\:\:\:\:\mathrm{x}^{\mathrm{2011}} \:+\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2011}} }\:=\:? \\ $$

Answered by Rajpurohith last updated on 26/Jan/23

Answered by SEKRET last updated on 26/Jan/23

(x−1)(x^2 +x+1)=0 x^3 =1 x∙(x^3 )^(670) +(1/(x∙(x^3 )^(670) )) = x+(1/x) x^2 +x+1=0 → /x x+(1/x)= −1

$$\:\:\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} =\mathrm{1} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{x}}\centerdot\left(\boldsymbol{\mathrm{x}}^{\mathrm{3}} \right)^{\mathrm{670}} +\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}\centerdot\left(\boldsymbol{\mathrm{x}}^{\mathrm{3}} \right)^{\mathrm{670}} }\:=\:\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}} \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}=\mathrm{0}\:\:\:\:\:\:\:\rightarrow\:\:\:\:\:/\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}=\:−\mathrm{1}\: \\ $$

Answered by Rasheed.Sindhi last updated on 26/Jan/23

x^2 +x+1=0; x^(2011) +(1/x^(2011) )=? x^2 =−x−1 ⇒x^3 =−x^2 −x=−(−x−1)−x=1 ▶x^(2011) + (1/x^(2011) )=(x^3 )^(670) ∙x+(1/((x^3 )^(670) ∙x)) 1^(670) ∙x+(1/(1^(670) ∙x))=x+(1/x)=((x^2 +1)/x) =((−x−1+1)/x)=−1

$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0};\:\:\:\:\:\:{x}^{\mathrm{2011}} +\frac{\mathrm{1}}{{x}^{\mathrm{2011}} }=? \\ $$$${x}^{\mathrm{2}} =−{x}−\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{3}} =−{x}^{\mathrm{2}} −{x}=−\left(−{x}−\mathrm{1}\right)−{x}=\mathrm{1} \\ $$$$\: \\ $$$$\blacktriangleright\mathrm{x}^{\mathrm{2011}} \:+\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2011}} }=\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{670}} \centerdot\mathrm{x}+\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{670}} \centerdot\mathrm{x}} \\ $$$$\mathrm{1}^{\mathrm{670}} \centerdot{x}+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{670}} \centerdot{x}}={x}+\frac{\mathrm{1}}{{x}}=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}} \\ $$$$=\frac{−{x}−\mathrm{1}+\mathrm{1}}{{x}}=−\mathrm{1} \\ $$

Commented by Spillover last updated on 26/Jan/23

perfect

$${perfect} \\ $$

Commented by Rasheed.Sindhi last updated on 27/Jan/23

ThanX for compliment.

$$\mathcal{T}{han}\mathcal{X}\:{for}\:{compliment}. \\ $$

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