Question Number 138135 by Deepak1298 last updated on 10/Apr/21
$$\left({x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right)=\mathrm{4} \\ $$$${find}\:{x}. \\ $$
Answered by ajfour last updated on 10/Apr/21
$${let}\:{x}=\frac{\mathrm{4}}{{t}} \\ $$$$\frac{\mathrm{16}}{{t}^{\mathrm{2}} }+\frac{\mathrm{64}}{{t}^{\mathrm{3}} }=\mathrm{4} \\ $$$$\Rightarrow\:\frac{\mathrm{4}{t}+\mathrm{16}}{{t}^{\mathrm{3}} }=\mathrm{1} \\ $$$$\Rightarrow\:{t}^{\mathrm{3}} −\mathrm{4}{t}−\mathrm{16}=\mathrm{0} \\ $$$${If} \\ $$$${t}^{\mathrm{3}} +{pt}+{q}=\mathrm{0} \\ $$$${t}=\left(−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:+\left(−\frac{{q}}{\mathrm{2}}−\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${here}\:\:={p}=−\mathrm{4},\:{q}=−\mathrm{16} \\ $$$${so}\:\:{t}=\left(\mathrm{8}+\sqrt{\mathrm{64}−\frac{\mathrm{64}}{\mathrm{27}}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:+\left(\mathrm{8}−\sqrt{\mathrm{64}−\frac{\mathrm{64}}{\mathrm{27}}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${t}=\left(\mathrm{8}+\frac{\mathrm{8}\sqrt{\mathrm{26}}}{\mathrm{3}\sqrt{\mathrm{3}}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{8}−\frac{\mathrm{8}\sqrt{\mathrm{26}}}{\mathrm{3}\sqrt{\mathrm{3}}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${x}=\frac{\mathrm{4}}{{t}} \\ $$$${x}=\frac{\mathrm{4}}{\left(\mathrm{8}+\frac{\mathrm{8}\sqrt{\mathrm{26}}}{\mathrm{3}\sqrt{\mathrm{3}}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{8}−\frac{\mathrm{8}\sqrt{\mathrm{26}}}{\mathrm{3}\sqrt{\mathrm{3}}}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$${x}=\frac{\mathrm{2}}{\left(\mathrm{1}+\sqrt{\frac{\mathrm{26}}{\mathrm{27}}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{1}−\sqrt{\frac{\mathrm{26}}{\mathrm{27}}}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$${x}\approx\mathrm{1}.\mathrm{3146} \\ $$