Question Number 164341 by ajfour last updated on 16/Jan/22
$$\:\:{x}^{\mathrm{4}} ={x}^{\mathrm{2}} +{cx} \\ $$$$\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={cx}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${let}\:\:\:\:{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}={t}\:\:\:\Rightarrow \\ $$$$\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${now}\:\:{let}\:\:\:{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}={z}\:\:\:\Rightarrow \\ $$$$\left({z}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} ={c}^{\mathrm{4}} \left({z}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${z}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}={p}\:\:\Rightarrow\:\:\left({p}^{\mathrm{2}} −\frac{{c}^{\mathrm{4}} }{\mathrm{4}}\right)^{\mathrm{2}} ={c}^{\mathrm{8}} \left({p}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:{p}^{\mathrm{4}} −\frac{{c}^{\mathrm{4}} {p}^{\mathrm{2}} }{\mathrm{2}}−{c}^{\mathrm{8}} {p}+\frac{{c}^{\mathrm{8}} }{\mathrm{16}}−\frac{{c}^{\mathrm{10}} }{\mathrm{2}}=\mathrm{0} \\ $$$${p}^{\mathrm{4}} −{Ap}^{\mathrm{2}} −{Bp}−\lambda=\mathrm{0} \\ $$$$\left({p}^{\mathrm{2}} +{sp}+{h}\right)\left({p}^{\mathrm{2}} −{sp}−\frac{\lambda}{{h}}\right)=\mathrm{0} \\ $$$${h}−\frac{\lambda}{{h}}={s}^{\mathrm{2}} −{A} \\ $$$${s}\left({h}+\frac{\lambda}{{h}}\right)={B} \\ $$$$\frac{{B}^{\mathrm{2}} }{{s}^{\mathrm{2}} }−\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{2}} =\mathrm{4}\lambda \\ $$$$\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{3}} +{A}\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{2}} +\mathrm{4}\lambda\left({s}^{\mathrm{2}} −{A}\right) \\ $$$$\:\:\:\:\:\:\:\:\:+\mathrm{4}\lambda{A}−{B}^{\mathrm{2}} =\mathrm{0} \\ $$$${m}^{\mathrm{3}} +{Am}^{\mathrm{2}} +\mathrm{4}\lambda{m}+\left(\mathrm{4}\lambda{A}−{B}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${let}\:\:\:{m}={w}−\frac{{A}}{\mathrm{3}}\:\:\:\Rightarrow \\ $$$${w}^{\mathrm{3}} +\left(\mathrm{4}\lambda−\frac{{A}^{\mathrm{2}} }{\mathrm{3}}\right){w}+\frac{\mathrm{2}{A}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{8}\lambda{A}}{\mathrm{3}}−{B}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{4}\lambda−\frac{{A}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{2}{c}^{\mathrm{10}} −\frac{{c}^{\mathrm{8}} }{\mathrm{4}}−\frac{{c}^{\mathrm{8}} }{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{c}^{\mathrm{8}} \left({c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$
Commented by Tawa11 last updated on 16/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$