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Question Number 125136 by Snail last updated on 08/Dec/20

$$Supposea,b,carenonzerorealnumbers$$$$satisfying{(ab+bc+ca)}^3=abc{(a+b+c)}^3.$$$$Provdthata,b,cmustbetermsofaGeometric$$$$Progession$$$$$$

Commented by Snail last updated on 08/Dec/20

$$Pleasetrytosolvewithoutexpandingthecubic$$$$inaneasymethod$$

Commented by Dwaipayan Shikari last updated on 08/Dec/20

$${\left(\frac{ab+bc+ca}{a+b+c}\right)}^3=abc$$$${\left(\frac{a^{2}r+a^2{r}^3+a^2{r}^2}{a+ar+{ar}^2}\right)}^3=a^3{r}^3=abc$$

Commented by Snail last updated on 08/Dec/20

$$Thisisthorrowlywrong...becauseu{can}^{\prime }t$$$$assumetheyareinG.P.....uhavetoprovethat$$$$uhaveprovedthevice-versaresult......Try$$$$anotherway$$

Answered by Dwaipayan Shikari last updated on 08/Dec/20

$$b=ka$$$$c=ta$$$${(ab+bc+ca)}^3={(a^{2}k+a^{2}kt+a^{2}t)}^3=a^6{(k+t+kt)}^3$$$${(a+b+c)}^3=a^3(k+t+1)$$$${(k+t+kt)}^3=kt{(k+t+1)}^{3}k=\frac{p}{t}$$$$\Rightarrow {(1+\frac{kt-1}{k+t+1})}^3=kt$$$$\Rightarrow {(1+\frac{p-1}{p(t+\frac{1}{t})+1})}^3=p\Rightarrow \frac{p-1}{p(t+\frac{1}{t})+1}=\sqrt[3]{p}-1p=m^3$$$$\Rightarrow \frac{m^2+m+1}{m^3(t+\frac{1}{t})+1}=1\Rightarrow m^2+m=m^3(t+\frac{1}{t})\Rightarrow m+1=m^2(t+\frac{1}{t})$$$$t=\frac{m+1+\sqrt{{(m+1)}^2-4m^4}}{2m^2}k=p\left(\frac{m+1-\sqrt{{(m+1)}^2-4m^4}}{2m^2}\right)$$$$p=1$$$$b=a\frac{1}{\left(\frac{m+1+\sqrt{{(m+1)}^2-4m^4}}{2m^2}\right)},c=a\left(\left(\frac{m+1+\sqrt{{(m+1)}^2-4m^4}}{2m^2}\right)\right)$$$$$$

Commented by Snail last updated on 08/Dec/20

$$Quitegood$$

Commented by Snail last updated on 09/Dec/20

$$Hereisamistakein5thlinewhereindenominator$$$$uhavewritenp(t+\frac{1}{t})insteadof(\frac{p}{t}+t)$$

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