Question

please important i want to know the answer​

Answer 1

Step-by-step explanation:

$since \: abc =1 \\ c = \frac{1}{ab} \\ now \: \frac{1}{1 + a + {b}^{ - 1} } = \frac{1}{1 + a + \frac{1}{b } } \\ = \frac{1}{ \frac{b + a + 1}{b} } \\ = \frac{b}{b + a + 1}$

similarly

$\frac{1}{1 + b + {c}^{ - 1} } = \frac{c}{c + bc + 1} \\ = \frac{ \frac{1}{ab} }{ \frac{1}{ab} + b ( \frac{1}{ab} )+ 1 } \\ = \frac{ \frac{1}{ab} }{ \frac{1 + b + ab}{ab} } \\ = \frac{1}{1 + ab + b}$

and

$\frac{1}{1 + c + {a}^{ - 1} } = \frac{a}{a + ac + 1} \\ = \frac{a}{a + a( \frac{1}{ab} ) + 1} \\ = \frac{a}{ \frac{ab + 1 + b}{b} } \\ = \frac{ab}{1 + ab + b}$

now

$\frac{1}{1 + a + {b}^{ - 1} } + \frac{1}{1 + b + {c}^{ - 1} } + \frac{1}{1 + c+ {a}^{ - 1} } \\ =\frac{b}{b + ab + 1} + \frac{1}{1 + ab + b} + \frac{ab}{1 + ab + b} \\ = \frac{1 + ab + b}{1 + ab + b} \\ = 1 \\ \\ hence \: proved$