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Question Number 150895 by tabata last updated on 16/Aug/21

Commented by tabata last updated on 16/Aug/21

$f i n d (a . b . c)$
	Answered by amin96 last updated on 16/Aug/21
	${ ((a+b+c=k)),((abc=n)),((ab+bc+ac=m)) :}⇒ { ((a^2 +b^2 +c^2 +2(ab+ac+bc)=k^2 )),((abc=n)),(((ab)^2 +(ac)^2 +(bc)^2 +2abc(a+b+c)=m^2 )) :} { ((a+b+c=(1/a)+(1/b)+(1/c))),((a^2 +b^2 +c^2 =(1/a^2 )+(1/b^2 )+(1/c^2 ))) :} ⇒ { ((k=(m/n))),((k^2 −2m=((m^2 −2nk)/n^(2 ) ))) :} ((m/n))^2 −2m=((m^2 −2m)/n^2 ) ⇒ m^2 −2mn^2 =m^2 −2m 2mn^2 =2m n=±1 ⇒ abc=±1 (a=±1 b=±1 c=±1)$
	${\begin{cases} a + b + c = k \\ a b c = n \\ a b + b c + a c = m \end{cases} \Rightarrow {\begin{cases} a^{2} + b^{2} + c^{2} + 2 (a b + a c + b c) = k^{2} \\ a b c = n \\ {(a b)}^{2} + {(a c)}^{2} + {(b c)}^{2} + 2 a b c (a + b + c) = m^{2} \end{cases}$ ${\begin{cases} a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \\ a^{2} + b^{2} + c^{2} = \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} \end{cases} \Rightarrow {\begin{cases} k = \frac{m}{n} \\ k^{2} - 2 m = \frac{m^{2} - 2 n k}{n^{2}} \end{cases}$ ${(\frac{m}{n})}^{2} - 2 m = \frac{m^{2} - 2 m}{n^{2}} \Rightarrow m^{2} - 2 {m n}^{2} = m^{2} - 2 m$ $2 {m n}^{2} = 2 m n = \pm 1 \Rightarrow a b c = \pm 1 (a = \pm 1 b = \pm 1 c = \pm 1)$
	Commented by Rasheed.Sindhi last updated on 16/Aug/21

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