Question Number 160424 by cortano last updated on 29/Nov/21
Answered by MJS_new last updated on 29/Nov/21
$$\mathrm{let}\:{y}={px}\:\Rightarrow \\ $$$$\begin{cases}{{x}\left(\left({p}^{\mathrm{4}} −\mathrm{1}\right){x}^{\mathrm{3}} −{p}+\mathrm{2}\right)=\mathrm{0}}\\{\left({p}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} {x}^{\mathrm{6}} +\mathrm{3}=\mathrm{0}}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\:{x}_{\mathrm{1}} =\left(\frac{{p}−\mathrm{2}}{{p}^{\mathrm{4}} −\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} ;\:{x}_{\mathrm{2}} =\omega{x}_{\mathrm{1}} ;\:{x}_{\mathrm{3}} =\omega^{\mathrm{2}} {x}_{\mathrm{1}} ;\:\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${p}^{\mathrm{4}} −{p}^{\mathrm{3}} +\frac{\mathrm{9}}{\mathrm{4}}{p}^{\mathrm{2}} +{p}−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({p}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}{p}^{\mathrm{2}} +\mathrm{3}{p}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${p}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${p}_{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{3}^{\mathrm{1}/\mathrm{3}} −\mathrm{3}^{\mathrm{2}/\mathrm{3}} }{\mathrm{2}} \\ $$$${p}_{\mathrm{3}} =\frac{\mathrm{1}−\mathrm{3}^{\mathrm{1}/\mathrm{3}} +\mathrm{3}^{\mathrm{2}/\mathrm{3}} }{\mathrm{2}}+\frac{\mathrm{3}^{\mathrm{5}/\mathrm{6}} +\mathrm{3}^{\mathrm{7}/\mathrm{6}} }{\mathrm{4}}\mathrm{i} \\ $$$${p}_{\mathrm{4}} =\frac{\mathrm{1}−\mathrm{3}^{\mathrm{1}/\mathrm{3}} +\mathrm{3}^{\mathrm{2}/\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{3}^{\mathrm{5}/\mathrm{6}} +\mathrm{3}^{\mathrm{7}/\mathrm{6}} }{\mathrm{4}}\mathrm{i} \\ $$$$\mathrm{now}\:\mathrm{please}\:\mathrm{complete}\:\mathrm{it}\:\mathrm{yourself} \\ $$
Commented by mr W last updated on 29/Nov/21
$${perfect}! \\ $$
Commented by cortano last updated on 29/Nov/21
$$\mathrm{waw}…\mathrm{great} \\ $$