Question Number 227319 by Lara2440 last updated on 15/Jan/26
$${f}\left({z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }\right)={z} \\ $$$${f}\left({z}\right)=? \\ $$
Answered by breniam last updated on 14/Jan/26
$$\left(\forall{z}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\}\right)\left({f}\left({z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }\right)={z}\right)\Rightarrow\left(\forall\frac{\mathrm{1}}{{z}}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\}\right)\left({f}\left(\frac{\mathrm{1}}{{z}^{\mathrm{3}} }+{z}^{\mathrm{3}} \right)={f}\left({z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }\right)=\frac{\mathrm{1}}{{z}}\right) \\ $$$$\left(\forall{z}\in\mathbb{R}^{+} \backslash\left\{\mathrm{0}\right\}\right)\left(\frac{\mathrm{1}}{{z}}={f}\left({z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }\right)={z}\right) \\ $$$$\mathrm{Are}\:\mathrm{you}\:\mathrm{sure}\:\mathrm{there}\:\mathrm{wasn}'\mathrm{t}\:\mathrm{any}\:\mathrm{extra}\:\mathrm{condition} \\ $$$$\mathrm{e}.\mathrm{g}.\:{z}\in\left(\mathrm{0},\mathrm{1}\right)?\:\mathrm{The}\:\mathrm{conclusion}\:\mathrm{above}\:\mathrm{surely} \\ $$$$\mathrm{implies}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{function}\:\mathrm{meeting}\:\mathrm{condition}\:\mathrm{given}. \\ $$
Answered by Ghisom_ last updated on 14/Jan/26
$${t}={z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} } \\ $$$$\Rightarrow\:{z}^{\mathrm{3}} =\frac{{t}\pm\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{f}\left({t}\right)=\mathrm{2}^{−\mathrm{1}/\mathrm{3}} \omega^{{k}} \left({t}\pm\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\mathrm{with}\:\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\:\mathrm{and}\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$