Question Number 143011 by ajfour last updated on 08/Jun/21
$$\:\:{x}^{\mathrm{4}} +{bx}^{\mathrm{2}} +{cx}={s} \\ $$$${let}\:\:{x}^{\mathrm{2}} ={px}+{t} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{ptx}+{t}^{\mathrm{2}} +{bx}^{\mathrm{2}} +{cx}={s} \\ $$$$\Rightarrow\:\left({p}^{\mathrm{2}} +{b}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{pt}+{c}\right){x} \\ $$$$\:\:\:\:\:\:\:\:\:\:={s}−{t}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left({p}^{\mathrm{2}} +{b}\right)\left({px}+{t}\right)+\left(\mathrm{2}{pt}+{c}\right){x} \\ $$$$\:\:\:\:\:\:\:={s}−{t}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{p}\left({p}^{\mathrm{2}} +{b}\right)+\mathrm{2}{pt}+{c}=\mathrm{0} \\ $$$${and}\:\:\left({p}^{\mathrm{2}} +{b}\right){t}={s}−{t}^{\mathrm{2}} \\ $$$$\left(\frac{{s}}{{t}}−{t}−{b}\right)\left(\frac{{s}}{{t}}+{t}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left({A}−{b}\right)\left({A}^{\mathrm{2}} +\mathrm{4}{s}\right)={c}^{\mathrm{2}} \\ $$$$\Rightarrow\:{A}^{\mathrm{3}} −{bA}^{\mathrm{2}} +\mathrm{4}{sA}−\mathrm{4}{bs}−{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:\:{A}={z}+\frac{{b}}{\mathrm{3}}\:\Rightarrow \\ $$$${z}^{\mathrm{3}} +\left(\mathrm{4}{s}−\frac{{b}^{\mathrm{2}} }{\mathrm{3}}\right){z}−\left(\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{8}{bs}}{\mathrm{3}}+{c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${D}=\left(\frac{{b}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{4}{bs}}{\mathrm{3}}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{b}^{\mathrm{2}} }{\mathrm{9}}−\frac{\mathrm{4}{s}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$${If}\:\:{s}=\mathrm{0},\:{b}=−\mathrm{1},\:{c}\rightarrow−{c}\:\: \\ $$$${then}\:{D}=\left(−\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{3}} \\ $$$$… \\ $$