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Question Number 199135 by hardmath last updated on 28/Oct/23
a + (1/a) = 3 find: a^5 + (1/a^5 ) = ?
$$\mathrm{a}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:=\:\mathrm{3} \\ $$$$\mathrm{find}:\:\:\:\mathrm{a}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }\:\:=\:\:? \\ $$
Answered by som(math1967) last updated on 28/Oct/23
123
$$\mathrm{123} \\ $$
Answered by a.lgnaoui last updated on 28/Oct/23
3^5 =a^5 +(1/a^5 )+5(a^3 +(1/a^3 ))+10(a+(1/a)) =a^5 +(1/a^5 )+5(a+(1/a))[a^2 +(1/a^2 )+1]= =a^5 +(1/a^5 )+5(a+(1/a))[(a+(1/a))^2 −1] ⇒a^5 +(1/a^5 )=3^5 −5×24=243−120 a^5 +(1/a^5 )=123
$$\mathrm{3}^{\mathrm{5}} =\mathrm{a}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }+\mathrm{5}\left(\mathrm{a}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }\right)+\mathrm{10}\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right) \\ $$$$\:\:\:=\mathrm{a}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }+\mathrm{5}\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right)\left[\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\mathrm{1}\right]= \\ $$$$=\mathrm{a}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }+\mathrm{5}\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right)\left[\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right)^{\mathrm{2}} −\mathrm{1}\right] \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }=\mathrm{3}^{\mathrm{5}} −\mathrm{5}×\mathrm{24}=\mathrm{243}−\mathrm{120} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{a}}^{\mathrm{5}} +\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{5}} }=\mathrm{123} \\ $$$$ \\ $$
Commented by hardmath last updated on 28/Oct/23
thank you professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$
Answered by Rasheed.Sindhi last updated on 28/Oct/23
a + (1/a) = 3 ; a^5 + (1/a^5 ) = ? (a+(1/a))^2 =3^2 a^2 +(1/a^2 )=9−2=7 (a+(1/a))^3 =3^3 a^3 +(1/a^3 )=27−3(a+(1/a))=27−3(3)=18 (a^2 +(1/a^2 ))(a^3 +(1/a^3 ))=a^5 +(1/a^5 )+a+(1/a) (7)(18)=a^5 +(1/a^5 )+3 a^5 +(1/a^5 )=7×18−3=123
$$\mathrm{a}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:=\:\mathrm{3}\:;\:\mathrm{a}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }\:\:=\:\:? \\ $$$$\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }=\mathrm{9}−\mathrm{2}=\mathrm{7} \\ $$$$\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right)^{\mathrm{3}} =\mathrm{3}^{\mathrm{3}} \\ $$$$\mathrm{a}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }=\mathrm{27}−\mathrm{3}\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right)=\mathrm{27}−\mathrm{3}\left(\mathrm{3}\right)=\mathrm{18} \\ $$$$\left(\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\right)\left(\mathrm{a}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }\right)=\mathrm{a}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }+\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}} \\ $$$$\left(\mathrm{7}\right)\left(\mathrm{18}\right)=\mathrm{a}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }+\mathrm{3} \\ $$$$\mathrm{a}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }=\mathrm{7}×\mathrm{18}−\mathrm{3}=\mathrm{123} \\ $$
Answered by Rasheed.Sindhi last updated on 28/Oct/23
a + (1/a) = 3 ; a^5 + (1/a^5 ) = ? •(a+(1/a))^2 =3^2 =9 a^2 +(1/a^2 )=9−2=7 •(a^2 +(1/a^2 ))^2 =7^2 =49 a^4 +(1/a^4 )=49−2=47 •(a^2 +(1/a^2 ))(a+(1/a))=(7)(3)=21 a^3 +(1/a^3 )+a+(1/a)=21 a^3 +(1/a^3 )=21−(a+(1/a))=21−3=18 (a^4 +(1/a^4 ))(a+(1/a))=(47)(3) a^5 +(1/a^5 )+a^3 +(1/a^3 )=141 a^5 +(1/a^5 )+18=141 a^5 +(1/a^5 )=141−18=123
$$\mathrm{a}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:=\:\mathrm{3}\:;\:\mathrm{a}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }\:\:=\:\:? \\ $$$$\bullet\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} =\mathrm{9} \\ $$$$\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }=\mathrm{9}−\mathrm{2}=\mathrm{7} \\ $$$$\bullet\left(\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} =\mathrm{49} \\ $$$$\mathrm{a}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{4}} }=\mathrm{49}−\mathrm{2}=\mathrm{47} \\ $$$$\bullet\left(\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\right)\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right)=\left(\mathrm{7}\right)\left(\mathrm{3}\right)=\mathrm{21} \\ $$$$\:\:\mathrm{a}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }+\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}=\mathrm{21} \\ $$$$\:\:\mathrm{a}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }=\mathrm{21}−\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right)=\mathrm{21}−\mathrm{3}=\mathrm{18} \\ $$$$\left(\mathrm{a}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{4}} }\right)\left(\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}\right)=\left(\mathrm{47}\right)\left(\mathrm{3}\right) \\ $$$$\mathrm{a}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }+\mathrm{a}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }=\mathrm{141} \\ $$$$\mathrm{a}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }+\mathrm{18}=\mathrm{141} \\ $$$$\mathrm{a}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }=\mathrm{141}−\mathrm{18}=\mathrm{123} \\ $$
Commented by hardmath last updated on 28/Oct/23
thank you professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$
Answered by ajfour last updated on 28/Oct/23
a^2 +(1/a^2 )=7 p^n +q^n =(p+q){p^(n−1) −qp^(n−2) +.... ....+p(−q)^(n−2) +(−q)^(n−1) } hence a^5 +(1/a^5 )=(a+(1/a))(a^4 −a^2 +1−(1/a^2 )+(1/a^4 )) ⇒ Q=3(a^4 +(1/a^4 )−a^2 −(1/a^2 )+1) =3(47−7+1) = 3×41=123
$${a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\mathrm{7} \\ $$$${p}^{{n}} +{q}^{{n}} =\left({p}+{q}\right)\left\{{p}^{{n}−\mathrm{1}} −{qp}^{{n}−\mathrm{2}} +….\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….+{p}\left(−{q}\right)^{{n}−\mathrm{2}} +\left(−{q}\right)^{{n}−\mathrm{1}} \right\} \\ $$$${hence} \\ $$$${a}^{\mathrm{5}} +\frac{\mathrm{1}}{{a}^{\mathrm{5}} }=\left({a}+\frac{\mathrm{1}}{{a}}\right)\left({a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}^{\mathrm{4}} }\right) \\ $$$$\Rightarrow\:\:{Q}=\mathrm{3}\left({a}^{\mathrm{4}} +\frac{\mathrm{1}}{{a}^{\mathrm{4}} }−{a}^{\mathrm{2}} −\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\mathrm{1}\right) \\ $$$$\:\:=\mathrm{3}\left(\mathrm{47}−\mathrm{7}+\mathrm{1}\right) \\ $$$$\:\:=\:\mathrm{3}×\mathrm{41}=\mathrm{123} \\ $$
Answered by Rasheed.Sindhi last updated on 28/Oct/23
a + (1/a) = 3; a^5 + (1/a^5 ) = ? a + (1/a) = 3⇒ { (((1/a)=3−a^★ )),((a^2 =3a−1^(★★) )) :} ^★ (1/a)=3−a (1/a^2 )=(3−a)^2 =a^2 −6a+9 =(3a−1)−6a+9=−3a+8 (1/a^3 )=((−3a+8)/a)=−3+8((1/a)) =−3+8(3−a)=−8a+21 (1/a^4 )=−8+((21)/a)=−8+21(3−a) =−21a+55 (1/a^5 )=−21+((55)/a)=−21+55(3−a) =−21+165−55a=−55a+144 ^(★★) a^2 =3a−1 a^3 =3a^2 −a=3(3a−1)−a=8a−3 a^4 =8a^2 −3a=8(3a−1)−3a=21a−8 a^5 =21a^2 −8a=21(3a−1)−8a=55a−21 a^5 + (1/a^5 ) =(55a−21)+(−55a+144) =123
$$ \\ $$$$\mathrm{a}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:=\:\mathrm{3};\:\mathrm{a}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }\:\:=\:\:? \\ $$$$\mathrm{a}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:=\:\mathrm{3}\Rightarrow\begin{cases}{\frac{\mathrm{1}}{\mathrm{a}}=\mathrm{3}−\mathrm{a}^{\bigstar} }\\{\mathrm{a}^{\mathrm{2}} =\mathrm{3a}−\mathrm{1}^{\bigstar\bigstar} }\end{cases} \\ $$$$\:^{\bigstar} \:\frac{\mathrm{1}}{\mathrm{a}}=\mathrm{3}−\mathrm{a} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }=\left(\mathrm{3}−\mathrm{a}\right)^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} −\mathrm{6a}+\mathrm{9} \\ $$$$\:\:\:\:\:=\left(\mathrm{3a}−\mathrm{1}\right)−\mathrm{6a}+\mathrm{9}=−\mathrm{3a}+\mathrm{8} \\ $$$$\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{3}} }=\frac{−\mathrm{3}\boldsymbol{\mathrm{a}}+\mathrm{8}}{\boldsymbol{\mathrm{a}}}=−\mathrm{3}+\mathrm{8}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}\right) \\ $$$$\:\:\:\:\:\:=−\mathrm{3}+\mathrm{8}\left(\mathrm{3}−\mathrm{a}\right)=−\mathrm{8a}+\mathrm{21} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{4}} }=−\mathrm{8}+\frac{\mathrm{21}}{\mathrm{a}}=−\mathrm{8}+\mathrm{21}\left(\mathrm{3}−\mathrm{a}\right) \\ $$$$\:\:\:\:\:\:=−\mathrm{21a}+\mathrm{55} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }=−\mathrm{21}+\frac{\mathrm{55}}{\mathrm{a}}=−\mathrm{21}+\mathrm{55}\left(\mathrm{3}−\mathrm{a}\right) \\ $$$$\:\:\:\:\:=−\mathrm{21}+\mathrm{165}−\mathrm{55a}=−\mathrm{55a}+\mathrm{144} \\ $$$$\: \\ $$$$\:^{\bigstar\bigstar} \mathrm{a}^{\mathrm{2}} =\mathrm{3a}−\mathrm{1} \\ $$$$\mathrm{a}^{\mathrm{3}} =\mathrm{3a}^{\mathrm{2}} −\mathrm{a}=\mathrm{3}\left(\mathrm{3a}−\mathrm{1}\right)−\mathrm{a}=\mathrm{8a}−\mathrm{3} \\ $$$$\mathrm{a}^{\mathrm{4}} =\mathrm{8a}^{\mathrm{2}} −\mathrm{3a}=\mathrm{8}\left(\mathrm{3a}−\mathrm{1}\right)−\mathrm{3a}=\mathrm{21a}−\mathrm{8} \\ $$$$\mathrm{a}^{\mathrm{5}} =\mathrm{21a}^{\mathrm{2}} −\mathrm{8a}=\mathrm{21}\left(\mathrm{3a}−\mathrm{1}\right)−\mathrm{8a}=\mathrm{55a}−\mathrm{21} \\ $$$$ \\ $$$$\mathrm{a}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }\:\:=\left(\mathrm{55a}−\mathrm{21}\right)+\left(−\mathrm{55a}+\mathrm{144}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{123} \\ $$
Answered by Skabetix last updated on 28/Oct/23
a^2 −3a+1=0 △=b^2 −4ac=9−4=5>0 x_1 and x_2 =((3±(√)5)/2) a^5 +(1/a^5 )=((3±(√5))/2)+(1/((3±(√5))/2))=123
$${a}^{\mathrm{2}} −\mathrm{3}{a}+\mathrm{1}=\mathrm{0} \\ $$$$\bigtriangleup={b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{9}−\mathrm{4}=\mathrm{5}>\mathrm{0} \\ $$$${x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{3}\pm\sqrt{}\mathrm{5}}{\mathrm{2}} \\ $$$${a}^{\mathrm{5}} +\frac{\mathrm{1}}{{a}^{\mathrm{5}} }=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\mathrm{1}}{\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}}=\mathrm{123} \\ $$

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