Question Number 129687 by ajfour last updated on 17/Jan/21
$$\:\:\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$ $$\left({solve}\:{for}\:{x}\:\:{even}\:{if}\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$
Answered by ajfour last updated on 18/Jan/21
$${Just}\:{now}\:{I}\:{developed}\:{a}\:{new} \\ $$ $${approximate}\:{formula}: \\ $$ $${x}=−\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} +\sqrt{\mathrm{5}\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} +\frac{\mathrm{4}}{\mathrm{3}}}\: \\ $$ $${example}: \\ $$ $${y}=\left({X}−\mathrm{7}\right)\left({X}+\mathrm{2}\right)\left({X}+\mathrm{5}\right) \\ $$ $$\:\:={X}^{\mathrm{3}} −\mathrm{39}{X}−\mathrm{70} \\ $$ $${y}=\mathrm{0}\:\:\Rightarrow\:\:{X}^{\mathrm{3}} −\mathrm{39}{X}−\mathrm{70}=\mathrm{0} \\ $$ $${let}\:\:{x}=\frac{{X}}{\:\sqrt{\mathrm{39}}} \\ $$ $$\Rightarrow\:\:{x}^{\mathrm{3}} −{x}−\frac{\mathrm{70}}{\mathrm{39}\sqrt{\mathrm{39}}}\:=\mathrm{0} \\ $$ $$\:{here}\:\:{c}=\frac{\mathrm{70}}{\mathrm{39}\sqrt{\mathrm{39}}} \\ $$ $${x}=−\frac{\left(\mathrm{35}\right)^{\mathrm{1}/\mathrm{3}} }{\:\sqrt{\mathrm{39}}}+\sqrt{\mathrm{5}×\frac{\left(\mathrm{35}\right)^{\mathrm{2}/\mathrm{3}} }{\mathrm{39}}+\frac{\mathrm{4}}{\mathrm{3}}} \\ $$ $${X}=\sqrt{\mathrm{39}}{x} \\ $$ $${X}=−\left(\mathrm{35}\right)^{\mathrm{1}/\mathrm{3}} +\sqrt{\mathrm{5}\left(\mathrm{35}\right)^{\mathrm{2}/\mathrm{3}} +\frac{\mathrm{4}×\mathrm{39}}{\mathrm{3}}} \\ $$ $${X}=\mathrm{7}.\mathrm{00022} \\ $$ $$\left(\mathcal{E}{ureka}\:!\right) \\ $$
Commented byDwaipayan Shikari last updated on 18/Jan/21
$${How}\:{did}\:{you}\:{derive}\:{it}\:{sir}? \\ $$