Question Number 216739 by ArshadS last updated on 17/Feb/25
$${Prove}\:{that}\:\:^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\:−^{\mathrm{3}} \sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}\:=\mathrm{1} \\ $$$${Question}#\mathrm{216694}\:{reposted}\:{for}\:{new}\:{answers} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Feb/25
$$\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:+\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:=\mathrm{1} \\ $$$${let}\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:+\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:−\mathrm{2}}\:={x} \\ $$$${And}\:\:\:{a}=\sqrt{\mathrm{5}}\:+\mathrm{2}\:,\:{b}=\sqrt{\mathrm{5}}\:−\mathrm{2} \\ $$$$\Rightarrow{a}−{b}=\mathrm{4}\:,\:{ab}=\mathrm{1} \\ $$$${a}−{b}=\left(\sqrt[{\mathrm{3}}]{{a}}\:\right)^{\mathrm{3}} −\left(\sqrt[{\mathrm{3}}]{{b}\:}\:\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\sqrt[{\mathrm{3}}]{{a}}\:−\sqrt[{\mathrm{3}}]{{b}}\:\right)\left(\left(\sqrt[{\mathrm{3}}]{{a}}\:\right)^{\mathrm{2}} +\left(\sqrt[{\mathrm{3}}]{{b}}\:\right)^{\mathrm{2}} +\left(\sqrt[{\mathrm{3}}]{{a}}\:\right)\left(\sqrt[{\mathrm{3}}]{{b}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\sqrt[{\mathrm{3}}]{{a}}\:−\sqrt[{\mathrm{3}}]{{b}}\:\right)\left(\left(\sqrt[{\mathrm{3}}]{{a}}\:−\sqrt[{\mathrm{3}}]{{b}}\:\right)^{\mathrm{2}} +\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{a}}\:\right)\left(\sqrt[{\mathrm{3}}]{{b}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4}=\left({x}\right)\left(\:\left({x}\right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{1}\right)\:\right) \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1}\:{or}\:{x}^{\mathrm{2}} +{x}+\mathrm{4}=\mathrm{0}\Rightarrow{x}\notin\mathbb{R} \\ $$$$\:{proved} \\ $$