Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 187858 by normans last updated on 23/Feb/23

Answered by som(math1967) last updated on 23/Feb/23

(1/a)+(1/b)+(1/c)=(1/(a+b+c)) ⇒(a+b+c)(ab+bc+ca)−abc=0 ⇒a^2 b+abc+a^2 c+ab^2 +b^2 c+abc+abc +bc^2 +c^2 a−abc=0 ⇒ab(a+c)+b^2 (a+c)+ca(a+c)+bc(a+c)=0 (a+c)(ab+b^2 +ca+bc)=0 (a+c){b(a+b)+c(a+b)}=0 ⇒(a+c)(a+b)(b+c)=0 if (a+c)=0∴a=−c (1/a^5 )+(1/b^5 )+(1/c^5 )=−(1/c^5 )+(1/b^5 )+(1/c^5 )=(1/b^5 ) again (1/((a+b+c)^5 ))=(1/b^5 ) if a+b=0⇒a=−b orb=−c gives (1/a^5 )+(1/b^5 )+(1/c^5 )=(1/((a+b+c)^5 ))

$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{a}+{b}+{c}} \\ $$$$\Rightarrow\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)−{abc}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} {b}+{abc}+{a}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}+{abc}+\cancel{{abc}} \\ $$$$\:+{bc}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}−\cancel{{abc}}=\mathrm{0} \\ $$$$\Rightarrow{ab}\left({a}+{c}\right)+{b}^{\mathrm{2}} \left({a}+{c}\right)+{ca}\left({a}+{c}\right)+{bc}\left({a}+{c}\right)=\mathrm{0} \\ $$$$\left({a}+{c}\right)\left({ab}+{b}^{\mathrm{2}} +{ca}+{bc}\right)=\mathrm{0} \\ $$$$\left({a}+{c}\right)\left\{{b}\left({a}+{b}\right)+{c}\left({a}+{b}\right)\right\}=\mathrm{0} \\ $$$$\Rightarrow\left({a}+{c}\right)\left({a}+{b}\right)\left({b}+{c}\right)=\mathrm{0} \\ $$$${if}\:\left({a}+{c}\right)=\mathrm{0}\therefore{a}=−{c} \\ $$$$\:\frac{\mathrm{1}}{{a}^{\mathrm{5}} }+\frac{\mathrm{1}}{{b}^{\mathrm{5}} }+\frac{\mathrm{1}}{{c}^{\mathrm{5}} }=−\frac{\mathrm{1}}{{c}^{\mathrm{5}} }+\frac{\mathrm{1}}{{b}^{\mathrm{5}} }+\frac{\mathrm{1}}{{c}^{\mathrm{5}} }=\frac{\mathrm{1}}{{b}^{\mathrm{5}} } \\ $$$${again}\:\frac{\mathrm{1}}{\left({a}+{b}+{c}\right)^{\mathrm{5}} }=\frac{\mathrm{1}}{{b}^{\mathrm{5}} } \\ $$$${if}\:{a}+{b}=\mathrm{0}\Rightarrow{a}=−{b}\:{orb}=−{c} \\ $$$${gives}\:\frac{\mathrm{1}}{{a}^{\mathrm{5}} }+\frac{\mathrm{1}}{{b}^{\mathrm{5}} }+\frac{\mathrm{1}}{{c}^{\mathrm{5}} }=\frac{\mathrm{1}}{\left({a}+{b}+{c}\right)^{\mathrm{5}} } \\ $$$$ \\ $$

Commented by normans last updated on 23/Feb/23

very nice your solution Sir

$${very}\:{nice}\:{your}\:{solution}\:{Sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com