Question Number 80881 by ajfour last updated on 07/Feb/20
$${v}=−\frac{\left(\mathrm{2}{b}+\mathrm{3}{cp}\right){p}}{\mathrm{3}+{bp}^{\mathrm{2}} } \\ $$ $$\:\frac{\mathrm{3}{v}^{\mathrm{2}} +\mathrm{2}{bpv}+\mathrm{3}{cp}+{b}}{\mathrm{1}+{bp}^{\mathrm{2}} +{cp}^{\mathrm{3}} }\:>\:\mathrm{0}\:\: \\ $$ $${b},{c}\:\in\:\mathbb{R}\:,\:{b}<\mathrm{0} \\ $$ $${Any}\:{non}-{zero}\:{real}\:{value}\:{of}\:{p} \\ $$ $${in}\:{terms}\:{of}\:{b},{c}\:\:{obeying}\:{above} \\ $$ $${condition}? \\ $$
Answered by ajfour last updated on 07/Feb/20
$${let}\:\:{p}=−\frac{{b}}{{c}} \\ $$ $${v}=\frac{−{b}^{\mathrm{2}} }{{c}\left(\mathrm{3}+\frac{{b}^{\mathrm{3}} }{{c}^{\mathrm{2}} }\right)}=−\frac{{b}^{\mathrm{2}} {c}}{\mathrm{3}{c}^{\mathrm{2}} +{b}^{\mathrm{3}} } \\ $$ $${S}=\left(\frac{{b}^{\mathrm{2}} {c}}{\mathrm{3}{c}^{\mathrm{2}} +{b}^{\mathrm{3}} }\right)^{\mathrm{2}} +\frac{\mathrm{2}{b}^{\mathrm{2}} }{{c}}\left(\frac{{b}^{\mathrm{2}} {c}}{\mathrm{3}{c}^{\mathrm{2}} +{b}^{\mathrm{3}} }\right)−\mathrm{2}{b} \\ $$ $${let}\:{me}\:{test}\:{it} \\ $$ $${x}^{\mathrm{3}} −\mathrm{49}{x}+\mathrm{120}=\mathrm{0} \\ $$ $${let}\:{x}=\frac{{u}+{v}}{\mathrm{1}+{pu}} \\ $$ $${u}^{\mathrm{3}} +{Su}+\frac{{v}^{\mathrm{3}} +{bv}+{c}}{\mathrm{1}+{bp}^{\mathrm{2}} +{cp}^{\mathrm{3}} }=\mathrm{0} \\ $$ $$\Rightarrow\:{u}^{\mathrm{3}} +{Su}+{T}=\mathrm{0} \\ $$ $${D}=\:\frac{{T}^{\:\:\mathrm{2}} }{\mathrm{4}}+\frac{{S}^{\mathrm{3}} }{\mathrm{27}} \\ $$ $${u}=\left\{−\frac{{T}}{\mathrm{2}}+\sqrt{{D}}\right\}^{\mathrm{1}/\mathrm{3}} +\left\{−\frac{{T}}{\mathrm{2}}−\sqrt{{D}}\right\}^{\mathrm{1}/\mathrm{3}} \\ $$ $$........ \\ $$