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Question Number 54607 by ajfour last updated on 07/Feb/19

If ((u^5 +v^5 )/((u+v)^5 )) = −(1/5) , find ((u^3 +v^3 )/((u+v)^3 )) = ?

$${If}\:\:\frac{{u}^{\mathrm{5}} +{v}^{\mathrm{5}} }{\left({u}+{v}\right)^{\mathrm{5}} }\:=\:−\frac{\mathrm{1}}{\mathrm{5}}\:, \\ $$$${find}\:\:\:\frac{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} }{\left({u}+{v}\right)^{\mathrm{3}} }\:\:=\:? \\ $$

Commented by MJS last updated on 08/Feb/19

I think it can be −(4/5) or −(1/5)

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:−\frac{\mathrm{4}}{\mathrm{5}}\:\mathrm{or}\:−\frac{\mathrm{1}}{\mathrm{5}} \\ $$

Commented by ajfour last updated on 08/Feb/19

yes sir, i get −(1/5).

$${yes}\:{sir},\:{i}\:{get}\:−\frac{\mathrm{1}}{\mathrm{5}}. \\ $$

Answered by mr W last updated on 08/Feb/19

let (v/u)=t ((u^5 +v^5 )/((u+v)^5 ))=((1+t^5 )/((1+t)^5 ))=((1−t+t^2 −t^3 +t^4 )/((1+t)^4 ))=−(1/5) 6−t+11t^2 −t^3 +6t^4 =0 (2t^2 −t+2)(3t^2 +t+3)=0 ⇒t^2 −(1/2)t+1=0 ...(i) ⇒t^2 +(1/3)t+1=0 ...(ii) let ((u^3 +v^3 )/((u+v)^3 ))=((1+t^3 )/((1+t)^3 ))=((1−t+t^2 )/((1+t)^2 ))=(1/s) s−st+st^2 =1+2t+t^2 ⇒t^2 +((2+s)/(1−s))t+1=0 with ((2+s)/(1−s))=−(1/2) ⇒s+5=0⇒s=−5 with ((2+s)/(1−s))=(1/3) ⇒4s+5=0⇒s=−(5/4) ⇒((u^3 +v^3 )/((u+v)^3 ))=(1/s)=−(1/5) or −(4/5)

$${let}\:\frac{{v}}{{u}}={t} \\ $$$$\frac{{u}^{\mathrm{5}} +{v}^{\mathrm{5}} }{\left({u}+{v}\right)^{\mathrm{5}} }=\frac{\mathrm{1}+{t}^{\mathrm{5}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{5}} }=\frac{\mathrm{1}−{t}+{t}^{\mathrm{2}} −{t}^{\mathrm{3}} +{t}^{\mathrm{4}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{4}} }=−\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{6}−{t}+\mathrm{11}{t}^{\mathrm{2}} −{t}^{\mathrm{3}} +\mathrm{6}{t}^{\mathrm{4}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{t}^{\mathrm{2}} −{t}+\mathrm{2}\right)\left(\mathrm{3}{t}^{\mathrm{2}} +{t}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{t}+\mathrm{1}=\mathrm{0}\:\:...\left({i}\right) \\ $$$$\Rightarrow{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{t}+\mathrm{1}=\mathrm{0}\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${let}\:\frac{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} }{\left({u}+{v}\right)^{\mathrm{3}} }=\frac{\mathrm{1}+{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{3}} }=\frac{\mathrm{1}−{t}+{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{s}} \\ $$$${s}−{st}+{st}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{t}+{t}^{\mathrm{2}} \\ $$$$\Rightarrow{t}^{\mathrm{2}} +\frac{\mathrm{2}+{s}}{\mathrm{1}−{s}}{t}+\mathrm{1}=\mathrm{0} \\ $$$${with}\:\frac{\mathrm{2}+{s}}{\mathrm{1}−{s}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{s}+\mathrm{5}=\mathrm{0}\Rightarrow{s}=−\mathrm{5} \\ $$$$ \\ $$$${with}\:\frac{\mathrm{2}+{s}}{\mathrm{1}−{s}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{4}{s}+\mathrm{5}=\mathrm{0}\Rightarrow{s}=−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\Rightarrow\frac{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} }{\left({u}+{v}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{{s}}=−\frac{\mathrm{1}}{\mathrm{5}}\:{or}\:−\frac{\mathrm{4}}{\mathrm{5}} \\ $$

Commented by ajfour last updated on 08/Feb/19

Thanks Sir, you have done better.

$${Thanks}\:{Sir},\:{you}\:{have}\:{done}\:{better}. \\ $$$$ \\ $$

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