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Question Number 176676 by Rasheed.Sindhi last updated on 24/Sep/22

If x^3 +(1/x^3 )=1, prove that x^5 +(1/x^5 )=−(x^4 +(1/x^4 ))

$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{1}, \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }=−\left(\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right) \\ $$

Answered by mr W last updated on 24/Sep/22

x+(1/x)=p x^2 +(1/x^2 )+2=p^2 x^2 +(1/x^2 )=p^2 −2 (x^2 +(1/x^2 ))(x+(1/x))=(p^2 −2)p x^3 +(1/x^3 )+x+(1/x)=(p^2 −2)p 1+p=(p^2 −2)p ⇒p^3 −3p−1=0 (x^2 +(1/x^2 ))^2 =(p^2 −2)^2 x^4 +(1/x^4 )+2=p^4 −4p^2 +4 ⇒x^4 +(1/x^4 )=p^4 −4p^2 +2 =p(3p+1)−4p^2 +2 =−p^2 +p+2 (x^3 +(1/x^3 ))(x^2 +(1/x^2 ))=p^2 −2 x^5 +(1/x^5 )+x+(1/x)=p^2 −2 ⇒x^5 +(1/x^5 )=p^2 −p−2=−(x^4 +(1/x^4 )) ✓

$${x}+\frac{\mathrm{1}}{{x}}={p} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}={p}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={p}^{\mathrm{2}} −\mathrm{2} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\left({x}+\frac{\mathrm{1}}{{x}}\right)=\left({p}^{\mathrm{2}} −\mathrm{2}\right){p} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}+\frac{\mathrm{1}}{{x}}=\left({p}^{\mathrm{2}} −\mathrm{2}\right){p} \\ $$$$\mathrm{1}+{p}=\left({p}^{\mathrm{2}} −\mathrm{2}\right){p} \\ $$$$\Rightarrow{p}^{\mathrm{3}} −\mathrm{3}{p}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\left({p}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2}={p}^{\mathrm{4}} −\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }={p}^{\mathrm{4}} −\mathrm{4}{p}^{\mathrm{2}} +\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={p}\left(\mathrm{3}{p}+\mathrm{1}\right)−\mathrm{4}{p}^{\mathrm{2}} +\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−{p}^{\mathrm{2}} +{p}+\mathrm{2} \\ $$$$\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)={p}^{\mathrm{2}} −\mathrm{2} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+{x}+\frac{\mathrm{1}}{{x}}={p}^{\mathrm{2}} −\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }={p}^{\mathrm{2}} −{p}−\mathrm{2}=−\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)\:\checkmark \\ $$

Commented by Rasheed.Sindhi last updated on 25/Sep/22

Successful Approach as always. ThanX a lot sir!

$$\mathcal{S}{uccessful}\:\mathcal{A}{pproach}\:{as}\:{always}. \\ $$$$\mathcal{T}{han}\mathcal{X}\:{a}\:{lot}\:\boldsymbol{{sir}}! \\ $$

Commented by Tawa11 last updated on 25/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 24/Sep/22

p_n =x^n +(1/x^n ) p_1 =e_1 p_2 =e_1 p_1 −2e_2 =e_1 ^2 −2 p_3 =e_1 p_2 −e_2 p_1 =e_1 ^3 −3e_1 =1 ⇒e_1 ^3 =1+3e_1 p_4 =e_1 p_3 −e_2 p_2 =e_1 −e_1 ^2 +2 p_5 =e_1 p_4 −e_2 p_3 =e_1 ^2 −e_1 ^3 +2e_1 −1 =e_1 ^2 −1−3e_1 +2e_1 −1 =−e_1 +e_1 ^2 −2=−p_4 ✓

$${p}_{{n}} ={x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} } \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} ={e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{3}{e}_{\mathrm{1}} =\mathrm{1}\:\Rightarrow{e}_{\mathrm{1}} ^{\mathrm{3}} =\mathrm{1}+\mathrm{3}{e}_{\mathrm{1}} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} ={e}_{\mathrm{1}} −{e}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −{e}_{\mathrm{1}} ^{\mathrm{3}} +\mathrm{2}{e}_{\mathrm{1}} −\mathrm{1} \\ $$$$\:\:\:\:\:={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}−\mathrm{3}{e}_{\mathrm{1}} +\mathrm{2}{e}_{\mathrm{1}} −\mathrm{1} \\ $$$$\:\:\:\:\:=−{e}_{\mathrm{1}} +{e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}=−{p}_{\mathrm{4}} \:\checkmark \\ $$

Commented by Rasheed.Sindhi last updated on 25/Sep/22

Successful Approach as always. ThanX a lot sir!

$$\mathcal{S}{uccessful}\:\mathcal{A}{pproach}\:{as}\:{always}. \\ $$$$\mathcal{T}{han}\mathcal{X}\:{a}\:{lot}\:\boldsymbol{{sir}}! \\ $$

Commented by Tawa11 last updated on 25/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Sep/22

AnOther way... x^6 =x^3 −1 x^(12) =(x^6 )^2 =(x^3 −1)^2 =x^6 −2x^3 +1 =x^3 −1−2x^3 +1=−x^3 x^5 +(1/x^5 )=(x^(12) /x^7 )+(x^7 /x^(12) )=((−x^3 )/x^7 )+(x^7 /(−x^3 )) =−(x^4 +(1/x^4 ))

$$\mathrm{AnOther}\:\mathrm{way}... \\ $$$${x}^{\mathrm{6}} ={x}^{\mathrm{3}} −\mathrm{1} \\ $$$${x}^{\mathrm{12}} =\left({x}^{\mathrm{6}} \right)^{\mathrm{2}} =\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{3}} −\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}=−{x}^{\mathrm{3}} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\frac{{x}^{\mathrm{12}} }{{x}^{\mathrm{7}} }+\frac{{x}^{\mathrm{7}} }{{x}^{\mathrm{12}} }=\frac{−{x}^{\mathrm{3}} }{{x}^{\mathrm{7}} }+\frac{{x}^{\mathrm{7}} }{−{x}^{\mathrm{3}} } \\ $$$$=−\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right) \\ $$

Commented by mr W last updated on 25/Sep/22

very smart!

$${very}\:{smart}! \\ $$

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