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Question Number 151174 by mathdanisur last updated on 18/Aug/21
((a+(√(3∙((a+(√(3∙((a+(√(3∙...))))^(1/3) ))))^(1/3) ))))^(1/3)   = 3   find  a=?
$$\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}\centerdot\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}\centerdot\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}\centerdot…}}}}}}\:\:=\:\mathrm{3}\: \\ $$$$\mathrm{find}\:\:{a}=? \\ $$
Answered by mr W last updated on 18/Aug/21
((a+(√(3×3))))^(1/3) =3  a+3=27  a=24
$$\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}×\mathrm{3}}}=\mathrm{3} \\ $$$${a}+\mathrm{3}=\mathrm{27} \\ $$$${a}=\mathrm{24} \\ $$
Commented by mathdanisur last updated on 18/Aug/21
Thankyou Ser
$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$
Answered by maged last updated on 18/Aug/21
let    ((a+(√(3∙((a+(√(3∙((a+(√(3∙...))))^(1/3) ))))^(1/3) ))))^(1/3)   = u    ⇒u=3  ((a+(√(3u))))^(1/3)  =3  a+(√(3u))=27  a+(√9)=27  a=27−3  a=24
$${let}\:\: \\ $$$$\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}\centerdot\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}\centerdot\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}\centerdot…}}}}}}\:\:=\:{u}\:\: \\ $$$$\Rightarrow{u}=\mathrm{3} \\ $$$$\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}{u}}}\:=\mathrm{3} \\ $$$${a}+\sqrt{\mathrm{3}{u}}=\mathrm{27} \\ $$$${a}+\sqrt{\mathrm{9}}=\mathrm{27} \\ $$$${a}=\mathrm{27}−\mathrm{3} \\ $$$${a}=\mathrm{24} \\ $$
Commented by mathdanisur last updated on 18/Aug/21
Thankyou Ser
$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$

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