Question-36061

Question Number 36061 by Raj Singh last updated on 28/May/18

Commented by Rasheed.Sindhi last updated on 28/May/18

N o t r e l a t e d t o t h e a b o v e q u e s t i o n .

D ear Abu behi,

I^{'} ve discovered relation between your

You can't use 'macro parameter character #' in math mode

And now I know that {you}^{'} ve a solution

You can't use 'macro parameter character #' in math mode

35844) . But wait \dots

Let Mr tanmay & S @ ty @ m complete

their trials .

BTW

D o you like to be called Abu behi by me ?

If not I will ever call you Mr Behi .

Commented by behi83417@gmail.com last updated on 28/May/18

d e a r m r R a s h e e d! y o u a r e r i g h t a n d

y o u a r e w e l c o m e .

Answered by math1967 last updated on 28/May/18

I t h i n k

L . H . S \frac{1}{1 + a + b^{- 1}} + \frac{1}{1 + b + c^{- 1}} + \frac{1}{1 + c + a^{- 1}}

= \frac{1}{1 + a + \frac{1}{b}} + \frac{1}{1 + b + \frac{1}{c}} + \frac{1}{1 + c + \frac{1}{a}}

= \frac{b}{b + a b + 1} + \frac{1}{1 + b + a b} + \frac{1}{1 + \frac{1}{a b} + \frac{1}{a}}

[∵ a b c = 1, ∴ \frac{1}{c} = a b a n d c = \frac{1}{a b}]

= \frac{b}{b + a b + 1} + \frac{1}{b + a b + 1} + \frac{a b}{a b + 1 + b}

= \frac{b + 1 + a b}{b + 1 + a b} = 1 = R . H . S

A m I c o r r e c t ?

Commented by Rasheed.Sindhi last updated on 28/May/18

e^{x} cellent!

Commented by math1967 last updated on 29/May/18

M y p l e a s u r e