similar-question-reposted-if-a-3-b-b-3-c-c-3-a-with-a-b-c-and-a-b-c-R-find-abc-2-

Question Number 161366 by mr W last updated on 17/Dec/21

[s i m i l a r q u e s t i o n r e p o s t e d]

i f a + \frac{3}{b} = b + \frac{3}{c} = c + \frac{3}{a} w i t h a \neq b \neq c

a n d a, b, c \in R . f i n d {(a b c)}^{2} = ?

Answered by 1549442205PVT last updated on 18/Dec/21

F r o m t h e h y p o t h e s i s w e h a v e :

a + \frac{3}{b} = c + \frac{3}{a} \Rightarrow c = \frac{a^{2} - 3}{a} + \frac{3}{b} (1) \Rightarrow \frac{3}{c} = \frac{3 a b}{a^{2} b + 3 a - 3 b}

\Rightarrow a + \frac{3}{b} = b + \frac{3 a b}{a^{2} b + 3 a - 3 b} \Rightarrow \frac{a b + 3 - b^{2}}{b} = \frac{3 a b}{a^{2} b + 3 a - 3 b}

\Leftrightarrow a^{3} b^{2} + 3 a^{2} b - a^{2} b^{3} + 3 a^{2} b + 9 a - 3 {a b}^{2} - 3 {a b}^{2} - 9 b + 3 b^{3} = 3 {a b}^{2}

A f t e r s i m p l i f y c a t i o n w e o b t a i n

a^{3} b^{2} - (b^{3} - 6 b) a^{2} - (9 b^{2} - 9) a + 3 b^{3} - 9 b

= (a - b) (a^{2} b^{2} + 6 a b - 3 b^{2} + 9) = 0 . S i n c e a \neq b

, w e g e t a^{2} b^{2} + 6 a b - 3 b^{2} + 9 = 0

\Leftrightarrow (a^{2} - 3) b^{2} + 6 a b + 9 = 0 .

∙ i f a^{2} - 3 = 0 \Leftrightarrow a = \pm \sqrt{3} t h e n b = \frac{- 3}{\pm 2 \sqrt{3}} = \frac{\mp \sqrt{3}}{2}

\Rightarrow c = \mp 2 \sqrt{3} \Rightarrow a b c = \pm 3 \sqrt{3} \Rightarrow {(a b c)}^{2} = 27

∙ i f a^{2} - 3 \neq 0 . T h e n △^{'} = 9 a^{2} - 9 a^{2} + 27 = 27, s o

b = \frac{- 3 a \pm 3 \sqrt{3}}{a^{2} - 3} . H e n c e f r o m (1) w e g e t

b c = 3 + \frac{a^{2} - 3}{a} . b = 3 + \frac{- 3 a \pm 3 \sqrt{3}}{a}

\Rightarrow a b c = 3 a - 3 a \pm 3 \sqrt{3} = \pm 3 \sqrt{3}

\Rightarrow {(a b c)}^{2} = 27

Commented by mr W last updated on 17/Dec/21

t h a n k s s i r!

Commented by 1549442205PVT last updated on 18/Dec/21

T h a n k y o u S i r M r . W, i m i s t a k e a n d c o r r e c t e d

Answered by mr W last updated on 17/Dec/21

a + \frac{3}{b} = b + \frac{3}{c} = c + \frac{3}{a} = k s a y

\Rightarrow b = \frac{3}{k - a}

\Rightarrow c = k - \frac{3}{a} = \frac{k a - 3}{a}

\frac{3}{k - a} + \frac{3 a}{k a - 3} = k

3 (k a - 3) + 3 a (k - a) = k (k - a) (k a - 3)

\Rightarrow (k^{2} - 3) (a^{2} - k a + 3) = 0

c a s e 1 : a^{2} - k a + 3 = 0

k = \frac{a^{2} + 3}{a} = a + \frac{3}{a}

\Rightarrow b = \frac{3}{k - a} = \frac{3}{a + \frac{3}{a} - a} = a

\Rightarrow c = k - \frac{3}{a} = a + \frac{3}{a} - \frac{3}{a} = a

t h a t m e a n s c a s e 1 i s n o t t r u e, s i n c e

i t l e a d s t o a = b = c .

c a s e 2 : k^{2} - 3 = 0 i . e . k = \pm \sqrt{3}

w i t h k = \sqrt{3}

\Rightarrow b = \frac{3}{k - a} = \frac{3}{\sqrt{3} - a} \neq a ✓

\Rightarrow c = \frac{k a - 3}{a} = \frac{\sqrt{3} a - 3}{a} = - \frac{\sqrt{3} (\sqrt{3} - a)}{a} \neq b \neq a ✓

\Rightarrow a b c = a \times \frac{3}{\sqrt{3} - a} \times (- \frac{\sqrt{3} (\sqrt{3} - a)}{a}) = - 3 \sqrt{3}

s i m i l a r l y w i t h k = - \sqrt{3}

\Rightarrow a b c = 3 \sqrt{3}

t h a t m e a n s w e a l w a y s h a v e

a b c = \pm 3 \sqrt{3}

i . e . {(a b c)}^{2} = 27 .

g e n e r a l l y

i f a + \frac{n}{b} = b + \frac{n}{c} = c + \frac{n}{a} a n d a \neq b \neq c,

t h e n {(a b c)}^{2} = n^{3} .

Commented by Tawa11 last updated on 17/Dec/21

Great sir .

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