Question Number 202447 by maths_plus last updated on 26/Dec/23
$$\mathrm{rationnalise}\:\mathrm{le}\:\mathrm{denominateur}\:\mathrm{de}\: \\ $$$$\mathrm{x}\:=\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{2}}−\mathrm{1}}{\mathrm{1}−^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}} \\ $$
Answered by cortano12 last updated on 27/Dec/23
$$\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} \:=\:\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right) \\ $$$$\:\mathrm{1}^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)^{\mathrm{3}} =\:\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{4}}\:\right) \\ $$$$\:\frac{\mathrm{1}}{\mathrm{3}}\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\right)\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{\mathrm{4}}\:\right)} \\ $$$$\:\Rightarrow\mathrm{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\:.\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}}{\mathrm{3}} \\ $$
Commented by Frix last updated on 27/Dec/23
$$\mathrm{Yes}\:\mathrm{but}\:{x}\neq\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}.\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}}{\mathrm{3}} \\ $$$${x}=\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}}{\mathrm{3}}=−\frac{\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{3}} \\ $$
Commented by cortano12 last updated on 27/Dec/23
$$\:\mathrm{why}? \\ $$
Commented by Frix last updated on 27/Dec/23
$$\mathrm{Sorry}\:\mathrm{there}\:\mathrm{was}\:\mathrm{a}\:\mathrm{typo}. \\ $$$$\mathrm{But}\:\mathrm{essentially}\:\mathrm{you}\:\mathrm{need}\:\mathrm{brackets}. \\ $$$${a}+{b}×\frac{{c}}{{d}}\neq\left({a}+{b}\right)×\frac{{c}}{{d}} \\ $$
Answered by MATHEMATICSAM last updated on 27/Dec/23
$$\mathrm{1}^{\mathrm{3}} \:+\:\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)^{\mathrm{3}} \:=\:\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{1}\right)\left(\mathrm{1}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right) \\ $$$$\mathrm{Now}\:{x}\:=\:\frac{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{1}\right)}{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{1}\right)\left(\mathrm{1}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)} \\ $$$$=\:\frac{\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)^{\mathrm{2}} \:−\:\mathrm{1}}{\mathrm{1}\:+\:\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)^{\mathrm{3}} }\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}\:−\:\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}}\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}\:−\:\mathrm{1}}{\mathrm{3}}\:\left(\mathrm{Ans}\right) \\ $$