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Question Number 226550 by gregori last updated on 03/Dec/25
a+b+c = x lim_(x→0) ((a^3 +b^3 +c^3 )/(abc)) =?
$$\:\:\:\: {a}+{b}+{c}\:=\:{x}\: \\ $$$$\:\:\:\: \underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{{abc}}\:=? \\ $$
Answered by peace2 last updated on 03/Dec/25
((a^3 +b^3 +c^3 )/(abc))≥((3abc)/(abc))=3 (a+b+c)^3 =a^3 +b^3 +3a^2 b+3b^2 a+3c^2 (a+b)+3c(a+b)^2 +c^3 =a^3 +b^3 +c^3 +3ab(a+b+c)+3c^2 b+3abc+3cb^2 +3ca^2 +3c^2 a =a^3 +b^3 +c^3 +3ab(a+b+c)+3cb(c+ab)+3ca(a+c) a^3 +b^3 +c^3 =^(x→0) −3ca(a+c)=−3ca(x−b)∼3abc ((3abc)/(abc))=3
$$\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{{abc}}\geqslant\frac{\mathrm{3}{abc}}{{abc}}=\mathrm{3} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{b}^{\mathrm{2}} {a}+\mathrm{3}{c}^{\mathrm{2}} \left({a}+{b}\right)+\mathrm{3}{c}\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{c}^{\mathrm{3}} \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}+{c}\right)+\mathrm{3c}^{\mathrm{2}} \mathrm{b}+\mathrm{3abc}+\mathrm{3cb}^{\mathrm{2}} +\mathrm{3ca}^{\mathrm{2}} +\mathrm{3c}^{\mathrm{2}} \mathrm{a} \\ $$$$=\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} +\mathrm{3ab}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)+\mathrm{3cb}\left(\mathrm{c}+\mathrm{ab}\right)+\mathrm{3ca}\left(\mathrm{a}+\mathrm{c}\right) \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \overset{\mathrm{x}\rightarrow\mathrm{0}} {=}−\mathrm{3}{ca}\left({a}+{c}\right)=−\mathrm{3}{ca}\left({x}−{b}\right)\sim\mathrm{3}{abc} \\ $$$$\frac{\mathrm{3}{abc}}{{abc}}=\mathrm{3} \\ $$$$ \\ $$$$ \\ $$
Answered by AgniMath last updated on 03/Dec/25
a^3 +b^3 +c^3 −3abc = (a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca) = x(a^2 +b^2 +c^2 −ab−bc−ca) ⇒ a^3 +b^3 +c^3 =3abc+x(a^2 +b^2 +c^2 −ab−bc−ca) lim_(x→0) ((a^3 +b^3 +c^3 )/(abc)) = lim_(x→0) [((3abc)/(abc))+((x(a^2 +b^2 +c^2 −ab−bc−ca))/(abc))] = 3+0 = 3
$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc} \\ $$$$=\:\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right) \\ $$$$=\:{x}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right) \\ $$$$\Rightarrow\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc}+{x}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{{abc}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{3}{abc}}{{abc}}+\frac{{x}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)}{{abc}}\right] \\ $$$$=\:\mathrm{3}+\mathrm{0}\:=\:\mathrm{3} \\ $$
Answered by Ghisom_ last updated on 03/Dec/25
a+b+c=0 ⇔ c=−(a+b) a^3 +b^3 +(−(a+b))^3 = a^3 +b^3 −(a^3 +3a^2 b+3ab^2 +b^3 )=−3ab(a+b) ab(−(a+b))=−ab(a+b) ((−3ab(a+b))/(−ab(a+b)))=3
$${a}+{b}+{c}=\mathrm{0}\:\Leftrightarrow\:{c}=−\left({a}+{b}\right) \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\left(−\left({a}+{b}\right)\right)^{\mathrm{3}} = \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} −\left({a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{ab}^{\mathrm{2}} +{b}^{\mathrm{3}} \right)=−\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$${ab}\left(−\left({a}+{b}\right)\right)=−{ab}\left({a}+{b}\right) \\ $$$$\frac{−\mathrm{3}{ab}\left({a}+{b}\right)}{−{ab}\left({a}+{b}\right)}=\mathrm{3} \\ $$

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