Question

If \(\quad \large abc = 1\), \(\quad\) then find:

\[\large \color{green}{(\frac{1}{1+a + b^{-1}} + \frac{1}{1 + b + c^{-1}} + \frac{1}{1 + c + a^{-1}})}\]

\[\large \color{blue}{\mathit{Answer : 1}}\]

Answer 1

so far i got..
\[\large \frac{6+3a+3b+3c+3bc+3ac+3ab+a^2b+b^2c+c^2a}{(b+ab+a)(c+bc+1)(a+ac+1)}\]
lol

Answer 2

Ha ha ha...
Try hard..

Answer 3

=( what do you suggest to do next?

Answer 4

Going totally wrong..
Even Einstein will fail to solve this what you are doing...

Answer 5

all denominators are equal,,thats what i got so far..

Answer 6

How you made all the denominator equal??

Answer 7

ohh sorry not equal at the moment,,we have this on simplification:

Answer 8

(a+b+c)/(1+(a-1) +c)

Answer 9

\[
\frac{1}{1+a+b^{-1}}\cdot\frac {bc} {bc}=\frac{bc}{bc+abc+c}=\frac{a^{-1}}{1+c+a^{-1}}
\]

Answer 10

no am sorry,,am getting confused..gimme 1 min..

Answer 11

After that...??

Answer 12

just let
a=x/y b=y/z c=z/x

Answer 13

Yes go ahead @mukushla ..

Answer 14

then u have
\[\frac{y}{x+y+z}+\frac{z}{x+y+z}+\frac{x}{x+y+z}=\frac{x+y+z}{x+y+z}=1\]

Answer 15

got it..1 min i'll post

Answer 16

I know one method to solve it..
Now I know two..

Thank you very much @mukushla

Answer 17

1/(1+a+(b-1)) = b/(b+(c-1)+1)
1/(1+c+(a-1) = a/(a+(b-1)+1) = ab/(b+(c-1)+1) = (c-1)/(b+(c-1)+1)
adding all
easily =1

Answer 18

If you go with @nphuongsun93 's attempt and expand those nasty multiplications, it does work out to be the same on top and bottom, = 1

It's just really hard to stay organized / avoid mistakes, and it takes a lot longer than the above methods. D:

Answer 19

I have another solution..
Everyone should know that:

Leave the first term as such, multiply and divide by \(b^{-1}\) to numerator and denominator of middle term and for third term, by taking LCD just solved the denominator,

You will get:

\[= \frac{1}{1 + a + b^{-1}} + \frac{b^{-1}}{b^{-1} + 1 + b^{-1}c^{-1}} + \frac{a}{a + ac + 1}\]

Using abc = 1

\[= \frac{1}{1 + a + b^{-1}} + \frac{b^{-1}}{1 + b^{-1} + a} + \frac{a}{a + b^{-1} + 1}\]

\[= \frac{1 + a + b^{-1}}{1 + a + b^{-1}} \implies = 1\]

Answer 20

hardest way for the hardworkers!!