Question Number 195765 by deleteduser4 last updated on 10/Aug/23
$${hello} \\ $$$$ \\ $$$$\:\begin{cases}{{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\mathrm{18}}\\{{x}>\mathrm{1}}\end{cases}\:\:\Rightarrow\:\:\:{x}^{\mathrm{5}} −\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\:=\:? \\ $$
Answered by mr W last updated on 10/Aug/23
![x^3 +(1/x^3 )=(x+(1/x))^3 −3(x+(1/x)) let s=x+(1/x) ∈R s^3 −3s−18=0 (s−3)(s^2 +3s+6)=0 ⇒s=3 (x^3 +(1/x^3 ))(x^2 +(1/x^2 ))=x^5 +(1/x^5 )+x+(1/x) x^5 +(1/x^5 )=(x^3 +(1/x^3 ))(x^2 +(1/x^2 ))−(x+(1/x)) x^5 +(1/x^5 )=(x^3 +(1/x^3 ))[(x+(1/x))^2 −2]−(x+(1/x)) x^5 +(1/x^5 )=18(s^2 −2)−s ⇒x^5 +(1/x^5 )=18(3^2 −2)−3=123 (x^5 −(1/x^5 ))^2 =(x^5 +(1/x^5 ))^2 −4 ⇒x^5 −(1/x^5 )=(√(123^2 −4))=55(√5) ✓](Q195769.png)
$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${let}\:{s}={x}+\frac{\mathrm{1}}{{x}}\:\in{R} \\ $$$${s}^{\mathrm{3}} −\mathrm{3}{s}−\mathrm{18}=\mathrm{0} \\ $$$$\left({s}−\mathrm{3}\right)\left({s}^{\mathrm{2}} +\mathrm{3}{s}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\Rightarrow{s}=\mathrm{3} \\ $$$$\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)={x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+{x}+\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)−\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\left[\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\right]−\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\mathrm{18}\left({s}^{\mathrm{2}} −\mathrm{2}\right)−{s} \\ $$$$\Rightarrow{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\mathrm{18}\left(\mathrm{3}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{3}=\mathrm{123} \\ $$$$\left({x}^{\mathrm{5}} −\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{2}} =\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{2}} −\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{5}} −\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\sqrt{\mathrm{123}^{\mathrm{2}} −\mathrm{4}}=\mathrm{55}\sqrt{\mathrm{5}}\:\checkmark \\ $$
Commented by deleteduser4 last updated on 10/Aug/23
$${Mr}.\:{W};\:{you}\:{are}\:{a}\:{genius}\:{in}\:{mathmatics} \\ $$$${thank}\:{you}\:{for}\:{nice}\:{your}\:{solution} \\ $$$$ \\ $$
Commented by mr W last updated on 10/Aug/23
$${i}'{m}\:{not}.\:{but}\:{thank}\:{you}\:{anyway}! \\ $$
Answered by ajfour last updated on 10/Aug/23
$${say}\:\:\:{x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}−\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:=\sqrt{\mathrm{324}−\mathrm{4}}=\mathrm{8}\sqrt{\mathrm{5}} \\ $$$$ \\ $$$${x}^{\mathrm{5}} −\frac{\mathrm{1}}{{x}^{\mathrm{5}} }={Q} \\ $$$$\:={Q}=\left({x}−\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}\right) \\ $$$${let}\:\:\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={t} \\ $$$$\frac{{Q}}{\mathrm{8}\sqrt{\mathrm{5}}}=\frac{{t}^{\mathrm{2}} −\mathrm{2}+{t}+\mathrm{1}}{{t}+\mathrm{1}} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{18}=\left({x}+\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}\right) \\ $$$$\mathrm{324}=\left({t}+\mathrm{2}\right)\left({t}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{t}=\mathrm{7} \\ $$$$\frac{{Q}}{\mathrm{8}\sqrt{\mathrm{5}}}=\frac{\mathrm{49}−\mathrm{2}+\mathrm{7}+\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow\:\:{Q}={x}^{\mathrm{5}} −\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\:\mathrm{55}\sqrt{\mathrm{5}} \\ $$
Commented by deleteduser4 last updated on 11/Aug/23
$${thank}\:{alot}\:{dear}\:{mr}\:{ajfour} \\ $$