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Question Number 173157 by AgniMath last updated on 07/Jul/22

If (1/a) + (1/b) + (1/c) = (1/(a + b + c)) then prove that (1/a^3 ) + (1/b^3 ) + (1/c^3 ) = (1/((a + b + c)^3 )) .

$$\mathrm{If}\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{{a}\:+\:{b}\:+\:{c}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{3}} }\:. \\ $$

Answered by Cesar1994 last updated on 07/Jul/22

(1/a)+(1/b)+(1/c)=(1/(a+b+c)) ((1/a)+(1/b)+(1/c))^3 =((1/(a+b+c)))^3 (1/a^3 )+(1/b^3 )+(1/c^3 )+3((1/a)+(1/b))((1/a)+(1/c))((1/b)+(1/c))=(1/((a+b+c)^3 )) ∙∙∙ (1) demostraremos que ((1/a)+(1/b))((1/a)+(1/c))((1/b)+(1/c))=0 Sea x=(1/(a+b+c)) ∙∙∙(2) ⇒((1/a)+(1/b))((1/a)+(1/c))((1/b)+(1/c))=(x−(1/c))(x−(1/b))(x−(1/a)) =x^3 −((1/a)+(1/b)+(1/c)_(x) )x^2 +((1/(ab))+(1/(bc))+(1/(ca)))x− (1/(abc)) =x^3 −x^3 +(((a+b+c)/(abc)))x−(1/(abc)) , (a+b+c)x=1 por (2) = (1/(abc))−(1/(abc)) =0 sustituyendo en (1) se concluye que (1/a^3 )+(1/b^3 )+(1/c^3 ) = (1/((a+b+c)^3 ))

$$ \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{a}+{b}+{c}} \\ $$$$\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)^{\mathrm{3}} =\left(\frac{\mathrm{1}}{{a}+{b}+{c}}\right)^{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} }+\frac{\mathrm{1}}{{b}^{\mathrm{3}} }+\frac{\mathrm{1}}{{c}^{\mathrm{3}} }+\mathrm{3}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}\right)\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)=\frac{\mathrm{1}}{\left({a}+{b}+{c}\right)^{\mathrm{3}} }\:\:\centerdot\centerdot\centerdot\:\left(\mathrm{1}\right) \\ $$$$\mathrm{demostraremos}\:\mathrm{que}\:\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}\right)\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)=\mathrm{0} \\ $$$$\mathrm{Sea}\:{x}=\frac{\mathrm{1}}{{a}+{b}+{c}}\:\centerdot\centerdot\centerdot\left(\mathrm{2}\right) \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}\right)\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)=\left({x}−\frac{\mathrm{1}}{{c}}\right)\left({x}−\frac{\mathrm{1}}{{b}}\right)\left({x}−\frac{\mathrm{1}}{{a}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{3}} −\left(\underset{{x}} {\underbrace{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}}}\right){x}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{ca}}\right){x}−\:\frac{\mathrm{1}}{{abc}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{3}} −{x}^{\mathrm{3}} +\left(\frac{{a}+{b}+{c}}{{abc}}\right){x}−\frac{\mathrm{1}}{{abc}}\:,\:\left({a}+{b}+{c}\right){x}=\mathrm{1}\:\mathrm{por}\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{{abc}}−\frac{\mathrm{1}}{{abc}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0} \\ $$$$\mathrm{sustituyendo}\:\mathrm{en}\:\left(\mathrm{1}\right)\:\mathrm{se}\:\mathrm{concluye}\:\mathrm{que} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} }+\frac{\mathrm{1}}{{b}^{\mathrm{3}} }+\frac{\mathrm{1}}{{c}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\left({a}+{b}+{c}\right)^{\mathrm{3}} } \\ $$

Commented by Tawa11 last updated on 11/Jul/22

Great sirs

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

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