Question Number 176676 by Rasheed.Sindhi last updated on 24/Sep/22
$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{1}, \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }=−\left(\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right) \\ $$
Answered by mr W last updated on 24/Sep/22
$${x}+\frac{\mathrm{1}}{{x}}={p} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}={p}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={p}^{\mathrm{2}} −\mathrm{2} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\left({x}+\frac{\mathrm{1}}{{x}}\right)=\left({p}^{\mathrm{2}} −\mathrm{2}\right){p} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}+\frac{\mathrm{1}}{{x}}=\left({p}^{\mathrm{2}} −\mathrm{2}\right){p} \\ $$$$\mathrm{1}+{p}=\left({p}^{\mathrm{2}} −\mathrm{2}\right){p} \\ $$$$\Rightarrow{p}^{\mathrm{3}} −\mathrm{3}{p}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\left({p}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2}={p}^{\mathrm{4}} −\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }={p}^{\mathrm{4}} −\mathrm{4}{p}^{\mathrm{2}} +\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={p}\left(\mathrm{3}{p}+\mathrm{1}\right)−\mathrm{4}{p}^{\mathrm{2}} +\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−{p}^{\mathrm{2}} +{p}+\mathrm{2} \\ $$$$\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)={p}^{\mathrm{2}} −\mathrm{2} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+{x}+\frac{\mathrm{1}}{{x}}={p}^{\mathrm{2}} −\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }={p}^{\mathrm{2}} −{p}−\mathrm{2}=−\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)\:\checkmark \\ $$
Commented by Rasheed.Sindhi last updated on 25/Sep/22
$$\mathcal{S}{uccessful}\:\mathcal{A}{pproach}\:{as}\:{always}. \\ $$$$\mathcal{T}{han}\mathcal{X}\:{a}\:{lot}\:\boldsymbol{{sir}}! \\ $$
Commented by Tawa11 last updated on 25/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 24/Sep/22
$${p}_{{n}} ={x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} } \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} ={e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{3}{e}_{\mathrm{1}} =\mathrm{1}\:\Rightarrow{e}_{\mathrm{1}} ^{\mathrm{3}} =\mathrm{1}+\mathrm{3}{e}_{\mathrm{1}} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} ={e}_{\mathrm{1}} −{e}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −{e}_{\mathrm{1}} ^{\mathrm{3}} +\mathrm{2}{e}_{\mathrm{1}} −\mathrm{1} \\ $$$$\:\:\:\:\:={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}−\mathrm{3}{e}_{\mathrm{1}} +\mathrm{2}{e}_{\mathrm{1}} −\mathrm{1} \\ $$$$\:\:\:\:\:=−{e}_{\mathrm{1}} +{e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}=−{p}_{\mathrm{4}} \:\checkmark \\ $$
Commented by Rasheed.Sindhi last updated on 25/Sep/22
$$\mathcal{S}{uccessful}\:\mathcal{A}{pproach}\:{as}\:{always}. \\ $$$$\mathcal{T}{han}\mathcal{X}\:{a}\:{lot}\:\boldsymbol{{sir}}! \\ $$
Commented by Tawa11 last updated on 25/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 25/Sep/22
$$\mathrm{AnOther}\:\mathrm{way}… \\ $$$${x}^{\mathrm{6}} ={x}^{\mathrm{3}} −\mathrm{1} \\ $$$${x}^{\mathrm{12}} =\left({x}^{\mathrm{6}} \right)^{\mathrm{2}} =\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{3}} −\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} +\mathrm{1}=−{x}^{\mathrm{3}} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\frac{{x}^{\mathrm{12}} }{{x}^{\mathrm{7}} }+\frac{{x}^{\mathrm{7}} }{{x}^{\mathrm{12}} }=\frac{−{x}^{\mathrm{3}} }{{x}^{\mathrm{7}} }+\frac{{x}^{\mathrm{7}} }{−{x}^{\mathrm{3}} } \\ $$$$=−\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right) \\ $$
Commented by mr W last updated on 25/Sep/22
$${very}\:{smart}! \\ $$