Question Number 54607 by ajfour last updated on 07/Feb/19
$${If}\:\:\frac{{u}^{\mathrm{5}} +{v}^{\mathrm{5}} }{\left({u}+{v}\right)^{\mathrm{5}} }\:=\:−\frac{\mathrm{1}}{\mathrm{5}}\:, \\ $$$${find}\:\:\:\frac{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} }{\left({u}+{v}\right)^{\mathrm{3}} }\:\:=\:? \\ $$
Commented by MJS last updated on 08/Feb/19
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:−\frac{\mathrm{4}}{\mathrm{5}}\:\mathrm{or}\:−\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by ajfour last updated on 08/Feb/19
$${yes}\:{sir},\:{i}\:{get}\:−\frac{\mathrm{1}}{\mathrm{5}}. \\ $$
Answered by mr W last updated on 08/Feb/19
$${let}\:\frac{{v}}{{u}}={t} \\ $$$$\frac{{u}^{\mathrm{5}} +{v}^{\mathrm{5}} }{\left({u}+{v}\right)^{\mathrm{5}} }=\frac{\mathrm{1}+{t}^{\mathrm{5}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{5}} }=\frac{\mathrm{1}−{t}+{t}^{\mathrm{2}} −{t}^{\mathrm{3}} +{t}^{\mathrm{4}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{4}} }=−\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{6}−{t}+\mathrm{11}{t}^{\mathrm{2}} −{t}^{\mathrm{3}} +\mathrm{6}{t}^{\mathrm{4}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{t}^{\mathrm{2}} −{t}+\mathrm{2}\right)\left(\mathrm{3}{t}^{\mathrm{2}} +{t}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{t}+\mathrm{1}=\mathrm{0}\:\:…\left({i}\right) \\ $$$$\Rightarrow{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{t}+\mathrm{1}=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${let}\:\frac{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} }{\left({u}+{v}\right)^{\mathrm{3}} }=\frac{\mathrm{1}+{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{3}} }=\frac{\mathrm{1}−{t}+{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{s}} \\ $$$${s}−{st}+{st}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{t}+{t}^{\mathrm{2}} \\ $$$$\Rightarrow{t}^{\mathrm{2}} +\frac{\mathrm{2}+{s}}{\mathrm{1}−{s}}{t}+\mathrm{1}=\mathrm{0} \\ $$$${with}\:\frac{\mathrm{2}+{s}}{\mathrm{1}−{s}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{s}+\mathrm{5}=\mathrm{0}\Rightarrow{s}=−\mathrm{5} \\ $$$$ \\ $$$${with}\:\frac{\mathrm{2}+{s}}{\mathrm{1}−{s}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{4}{s}+\mathrm{5}=\mathrm{0}\Rightarrow{s}=−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\Rightarrow\frac{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} }{\left({u}+{v}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{{s}}=−\frac{\mathrm{1}}{\mathrm{5}}\:{or}\:−\frac{\mathrm{4}}{\mathrm{5}} \\ $$
Commented by ajfour last updated on 08/Feb/19
$${Thanks}\:{Sir},\:{you}\:{have}\:{done}\:{better}. \\ $$$$ \\ $$