Question Number 86825 by M±th+et£s last updated on 31/Mar/20
$${a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=\mathrm{18} \\ $$$${a}^{\mathrm{4}} +\frac{\mathrm{1}}{{a}^{\mathrm{4}} }=? \\ $$
Commented by Ar Brandon last updated on 31/Mar/20
$${a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=\mathrm{18}\Rightarrow\:\:\:\left({a}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{18}{a}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$${a}^{\mathrm{3}} =\frac{\mathrm{18}\pm\sqrt{\left(−\mathrm{18}\right)^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\:\:\:=\:\:\:\frac{\mathrm{18}\pm\sqrt{\mathrm{320}}}{\mathrm{2}}\:\:=\mathrm{9}\pm\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$ \\ $$$${a}=\left(\mathrm{9}\pm\mathrm{4}\sqrt{\mathrm{5}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$ \\ $$$${a}^{\mathrm{4}} +\frac{\mathrm{1}}{{a}^{\mathrm{4}} }\:\:=\:\:\left(\mathrm{9}\pm\mathrm{4}\sqrt{\mathrm{5}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} \:+\:\frac{\mathrm{1}}{\left(\mathrm{9}\pm\mathrm{4}\sqrt{\mathrm{5}}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} } \\ $$
Answered by mr W last updated on 31/Mar/20
![(a+(1/a))^3 =a^3 +(1/a^3 )+3(a+(1/a))=18+3(a+(1/a)) let A=a+(1/a) A^3 −3A−18=0 (A−3)(A^2 +3A+6)=0 ⇒A=3=a+(1/a) a^4 +(1/a^4 )=(a^2 +(1/a^2 ))^2 −2=[(a+(1/a))^2 −2]^2 −2 =[3^2 −2]^2 −2 =47](https://www.tinkutara.com/question/Q86828.png)
$$\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }+\mathrm{3}\left({a}+\frac{\mathrm{1}}{{a}}\right)=\mathrm{18}+\mathrm{3}\left({a}+\frac{\mathrm{1}}{{a}}\right) \\ $$$${let}\:{A}={a}+\frac{\mathrm{1}}{{a}} \\ $$$${A}^{\mathrm{3}} −\mathrm{3}{A}−\mathrm{18}=\mathrm{0} \\ $$$$\left({A}−\mathrm{3}\right)\left({A}^{\mathrm{2}} +\mathrm{3}{A}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\Rightarrow{A}=\mathrm{3}={a}+\frac{\mathrm{1}}{{a}} \\ $$$${a}^{\mathrm{4}} +\frac{\mathrm{1}}{{a}^{\mathrm{4}} }=\left({a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{2}=\left[\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} −\mathrm{2}\right]^{\mathrm{2}} −\mathrm{2} \\ $$$$=\left[\mathrm{3}^{\mathrm{2}} −\mathrm{2}\right]^{\mathrm{2}} −\mathrm{2} \\ $$$$=\mathrm{47} \\ $$
Commented by M±th+et£s last updated on 31/Mar/20
$${thank}\:{you}\:{sir} \\ $$