Question Number 175866 by Linton last updated on 08/Sep/22
$${f}\left({x}\right)={x}^{\mathrm{3}} +{ax} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=? \\ $$
Answered by mr W last updated on 08/Sep/22
$${y}={x}^{\mathrm{3}} +{ax} \\ $$$${x}^{\mathrm{3}} +{ax}−{y}=\mathrm{0} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\sqrt{\frac{{y}^{\mathrm{2}} }{\mathrm{4}}+\frac{{a}^{\mathrm{3}} }{\mathrm{27}}}+\frac{{y}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{{y}^{\mathrm{2}} }{\mathrm{4}}+\frac{{a}^{\mathrm{3}} }{\mathrm{27}}}−\frac{{y}}{\mathrm{2}}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{a}^{\mathrm{3}} }{\mathrm{27}}}+\frac{{x}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{a}^{\mathrm{3}} }{\mathrm{27}}}−\frac{{x}}{\mathrm{2}}} \\ $$
Commented by peter frank last updated on 08/Sep/22
$$\mathrm{please}\:\mathrm{explain}\:\mathrm{3}^{\mathrm{rd}\:} \mathrm{line} \\ $$
Commented by mr W last updated on 09/Sep/22
$${cardano}'{s}\:{formula} \\ $$$${see}\:{Q}\mathrm{89687} \\ $$
Commented by Tawa11 last updated on 15/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$