Question Number 149883 by liberty last updated on 08/Aug/21
$$\mathrm{Prove}\:\mathrm{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{rational}\:\mathrm{number} \\ $$
Answered by john_santu last updated on 08/Aug/21
$$\mathrm{L}{et}\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\: \\ $$$${We}\:{then}\:{have}\:{x}−\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:=\mathrm{0} \\ $$$${as}\:{we}\:{have}\:{seen}\:{a}+{b}+{c}\:=\:\mathrm{0}\:{implies} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc}\:{so}\:{we}\:{obtain} \\ $$$${x}^{\mathrm{3}} −\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)−\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)=\mathrm{3}{x}\sqrt[{\mathrm{3}}]{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)} \\ $$$${or}\:{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}=\mathrm{0}\:.\:{clearly}\:{one}\:{of} \\ $$$${the}\:{roots}\:{of}\:{this}\:{equation}\:{is}\:{x}=\mathrm{1} \\ $$$${and}\:{the}\:{other}\:{two}\:{roots}\:{satisfy} \\ $$$${the}\:{equation}\:{x}^{\mathrm{2}} +{x}+\mathrm{4}=\mathrm{0}\:{which} \\ $$$${has}\:{no}\:{real}\:{solutions}.\:{since} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:{is}\:{a}\:{real}\:{root} \\ $$$${it}\:{follows}\:{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:=\mathrm{1} \\ $$$${which}\:{a}\:{rational}\:{number} \\ $$