Question Number 153226 by ajfour last updated on 05/Sep/21
Commented by mr W last updated on 05/Sep/21
$${x}^{\mathrm{3}} +{x}−{c}=\mathrm{0} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}+\frac{{c}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}−\frac{{c}}{\mathrm{2}}} \\ $$$${any}\:{other}\:{ways}\:{to}\:{solve}? \\ $$
Commented by ajfour last updated on 05/Sep/21
Commented by TVTA last updated on 06/Sep/21
$${x}^{\mathrm{3}} +{x}−{c}=\mathrm{0} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}+\frac{{c}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}−\frac{{c}}{\mathrm{2}}} \\ $$$${any}\:{other}\:{ways}\:{to}\:{solve}? \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{{c}+\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{27}}}}{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{{c}+\sqrt{{c}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{27}}}}{\mathrm{2}}}} \\ $$$${is}\:\:{this}\:{correct}\:{and}\:{ok}\:{sir}\:? \\ $$
Commented by ajfour last updated on 06/Sep/21
$${thanks}\:{everyone}! \\ $$
Answered by talminator2856791 last updated on 06/Sep/21
$$\:\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{2}{x}\:=\:{c} \\ $$$$\:\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{2}{x}\:=\:{c} \\ $$$$\:\mathrm{2}{x}^{\mathrm{3}} \:=\:{c} \\ $$$$\:{x}^{\mathrm{3}} \:=\:{c}\centerdot\mathrm{2}^{−\mathrm{1}} \\ $$$$\:{x}\:=\:\left({c}\centerdot\mathrm{2}^{−\mathrm{1}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$
Commented by MJS_new last updated on 06/Sep/21
$$\mathrm{formula}\:\mathrm{for}\:\mathrm{area}\:\mathrm{of}\:\mathrm{trapezoid}\:\mathrm{with}\:\mathrm{sides} \\ $$$${a}\parallel{c}\:\mathrm{and}\:{d}={b}\:\mathrm{with}\:\mathrm{given}\:\mathrm{height}\:{h}\:\mathrm{is}\:\frac{\left({a}+{c}\right){h}}{\mathrm{2}} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$$\frac{\left(\mathrm{2}+\mathrm{2}{x}^{\mathrm{2}} \right){x}}{\mathrm{2}}={c}\:\Leftrightarrow\:{x}^{\mathrm{3}} +{x}={c} \\ $$