Question Number 175095 by Gbenga last updated on 18/Aug/22
$$\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{c}}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$
Answered by MJS_new last updated on 18/Aug/22
$${x}={z}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{2}}{\mathrm{3}}{z}+\left({c}−\frac{\mathrm{7}}{\mathrm{27}}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${z}_{\mathrm{1}} ={u}+{v} \\ $$$${z}_{\mathrm{2}} =\omega{u}+\omega^{\mathrm{2}} {v} \\ $$$${z}_{\mathrm{3}} =\omega^{\mathrm{2}} {u}+\omega{v} \\ $$$$\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$${u}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}}}\wedge{v}=−\sqrt[{\mathrm{3}}]{\frac{{q}}{\mathrm{2}}\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}}} \\ $$$${p}=\frac{\mathrm{2}}{\mathrm{3}}\wedge{q}={c}−\frac{\mathrm{7}}{\mathrm{27}} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}=\frac{{c}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{7}{c}}{\mathrm{54}}+\frac{\mathrm{1}}{\mathrm{36}}>\mathrm{0}\forall{c}\in\mathbb{R} \\ $$$$\mathrm{do}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{yourself} \\ $$
Commented by Gbenga last updated on 19/Aug/22
$$ \\ $$
Commented by Tawa11 last updated on 19/Aug/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$