Question Number 187166 by ajfour last updated on 14/Feb/23
$${x}^{\mathrm{3}} ={x}+{c} \\ $$$${let}\:\:{x}=\frac{{mt}}{\mathrm{1}−{t}} \\ $$$${m}^{\mathrm{3}} {t}^{\mathrm{3}} ={mt}\left(\mathrm{1}−{t}\right)^{\mathrm{2}} +{c}\left(\mathrm{1}−{t}\right)^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\left({m}^{\mathrm{3}} −{m}−{c}\right){t}^{\mathrm{3}} +\left(\mathrm{2}{m}−\mathrm{3}{c}\right){t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:+\left(\mathrm{3}{c}−{m}\right){t}−{c}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} +{At}^{\mathrm{2}} +{Bt}+{C}=\mathrm{0} \\ $$$${let}\:\:{AB}={C} \\ $$$$\Rightarrow\:\left(\mathrm{2}{m}−\mathrm{3}{c}\right)\left({m}−\mathrm{3}{c}\right)={c}\left({m}^{\mathrm{3}} −{m}−{c}\right) \\ $$$$\Rightarrow\:{m}^{\mathrm{3}} −\frac{\mathrm{2}}{{c}}{m}^{\mathrm{2}} −\mathrm{8}{m}+\mathrm{10}{c}=\mathrm{0} \\ $$$$… \\ $$