Given { ((a^2 +ab+bc+ac=a+c)),((b^2 +ab+bc+ac=b+a)),((c^2 +ab+bc+ac=c+b)) :} find the value of a+b+c


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Question Number 109500 by john santu last updated on 24/Aug/20
$Given { ((a^2 +ab+bc+ac=a+c)),((b^2 +ab+bc+ac=b+a)),((c^2 +ab+bc+ac=c+b)) :} find the value of a+b+c$
$G i v e n {\begin{cases} a^{2} + a b + b c + a c = a + c \\ b^{2} + a b + b c + a c = b + a \\ c^{2} + a b + b c + a c = c + b \end{cases}$ $f i n d t h e v a l u e o f a + b + c$
	Answered by bemath last updated on 24/Aug/20

	$\Leftrightarrow {(a + b + c)}^{2} + a b + b c + a c = 2 (a + b + c)$ $l e t a + b + c = p$ $\Rightarrow p^{2} + a b + b c + a c = p$ $\Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{a b + a c + b c}{a b c}$
	Answered by mr W last updated on 24/Aug/20

	$d u e t o s y m m e t r y :$ $a = b = c$ $\Rightarrow 4 a^{2} = 2 a$ $\Rightarrow a = 0 o r a = \frac{1}{2}$ $\Rightarrow a + b + c = 0 o r \frac{3}{2}$
	Commented by bemath last updated on 24/Aug/20

	$s i r . h o w t o k n o w t h a t e q u a t i o n d u e t o$ $s y m e t r y s i r ?$
	Commented by mr W last updated on 24/Aug/20

	$i f y o u r e p l a c e a w i t h b o r c, b w i t h c o r$ $a, c w i t h a o r b, a n d t h e e q u a t i o n s$ $r e m a i n t h e s a m e, t h e n t h e y a r e$ $s y m m e t r i c . a = b = c c o u l d b e t h e$ $s o l u t i o n, b u t {m u s t n}^{'} t .$
	Commented by bemath last updated on 24/Aug/20

	$t h a n k y o u s i r$
	Answered by Rasheed.Sindhi last updated on 24/Aug/20

	$a + c - a^{2} = b + a - b^{2} = c + b - c^{2}$ $b + a - b^{2} - a - c + a^{2} = 0$ $(a - b) (a + b - 1) = 0 \times$ $a = b ∣ a + b = 1$ $S i m i l a r l y$ $b = c ∣ b + c = 1$ $c = a ∣ c + a = 1$ $a = b = c ∣ 2 (a + b + c) = 3$ $a + b + c = 3 k \forall k \in Z ∣ a + b + c = 3 / 2$
	Commented by mr W last updated on 24/Aug/20

	$i m e a n w h e n a = b = c = k, t h e n$ $4 k^{2} = 2 k$ $\Rightarrow k = 0 o r k = \frac{1}{2}$ $t h e r e i s n o o t h e r s o l u t i o n s .$
	Commented by mr W last updated on 24/Aug/20

	$k = 0 ?$
	Commented by Rasheed.Sindhi last updated on 24/Aug/20

	$Y e s s i r, I t h i n k a = b = c = 0 a l s o$ $s a t i s f y t h e g i v e n e q u a t i o n s .$
	Commented by $@y@m last updated on 24/Aug/20

	$@ M r . R a s h i d S i n d h i!$ $H o w c d i s a p p e a r e d i n 3^{r d} l i n e ?$
	Commented by $@y@m last updated on 24/Aug/20

	$@ M r . W!$ $S o r r y,$ $N o d o u b t o n y o u r s o l u t i o n .$ $T h i s a p p d o e s n o t s p e c i f y$ $c o m m e n t o v e r c o m m e n t . A l l g o e s$ $i n a c o n t i n u o u s c h a i n .$
	Commented by Rasheed.Sindhi last updated on 24/Aug/20

	$@ S @ y @ m$ $b - b^{2} - c + a^{2} = 0$ $a^{2} - b^{2} + b - c$ $(a - b) (a + b) + b - c$ $I t h o u g h t c a s a! T h i s m i s t a k e$ $m a d e m y c a l c u l a t i o n e a s y b u t$ $y o u c a u g h t m y m i s t a k e . h a h a h a . .$ $a n d m a k e m y w a y d i f f i c u l t!$
	Commented by Rasheed.Sindhi last updated on 24/Aug/20

	$@ M r W$ $s i r t h e s o l u t i o n i s w r o n g .$
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