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