Question

prove this please full step required ​

Answer 1

$\large\underline{\sf{Given- }}$

$\boxed{ \rm{ abc + 1 = 0}}$

$\large\underline{\sf{To\:prove - }}$

$\boxed{ \rm{ \frac{1}{1 - a - {b}^{ - 1} } + \frac{1}{1 - b - {c}^{ - 1} } + \frac{1}{1 - c - {a}^{ - 1} } = 1 }}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:abc + 1 = 0$

$\bf\implies \:abc = - 1 - - - (1)$

Consider,

$\rm :\longmapsto\:\dfrac{1}{1 - a - {b}^{ - 1} } + \dfrac{1}{1 - b - {c}^{ - 1} } + \dfrac{1}{1 - c - {a}^{ - 1} }$

Consider first term

$\rm :\longmapsto\:\dfrac{1}{1 - a - {b}^{ - 1} }$

$\rm \: = \: \: \dfrac{1}{1 - a - \dfrac{1}{b} }$

$\rm \: = \: \: \dfrac{b}{b - ab - 1}$

Now, Consider second term

$\rm :\longmapsto\: \dfrac{1}{1 - b - {c}^{ - 1} }$

$\rm \: = \: \: \dfrac{1}{1 - b - \dfrac{1}{c} }$

As it is given that,

$\boxed{ \rm{ abc = - 1\rm \: \implies\:ab = - \frac{1}{c}}}$

So, using this, we get

$\rm \: = \: \: \dfrac{1}{1 - b + ab }$

Now, Consider third term,

$\rm :\longmapsto\: \dfrac{1}{1 - c - {a}^{ - 1} }$

As it is given that

$\boxed{ \rm{ abc = - 1\rm \: \implies\:c = - \frac{1}{ab}}}$

$\rm \: = \: \: \dfrac{1}{1 + \dfrac{1}{ab} - \dfrac{1}{a} }$

$\rm \: = \: \: \dfrac{ab}{ab + 1 - b}$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{1 - a - {b}^{ - 1} } + \dfrac{1}{1 - b - {c}^{ - 1} } + \dfrac{1}{1 - c - {a}^{ - 1} }$

$\rm \: = \: \: \dfrac{b}{b - ab - 1} + \dfrac{1}{1 - b + ab} + \dfrac{ab}{ab + 1 - b}$

$\rm \: = \: \: \dfrac{b}{ - (1 - b + ab)} + \dfrac{1}{1 - b + ab} + \dfrac{ab}{1 - b + ab}$

$\rm \: = \: \: \dfrac{ - b}{1 - b + ab} + \dfrac{1}{1 - b + ab} + \dfrac{ab}{1 - b + ab}$

$\rm \: = \: \: \dfrac{ - b + 1 + ab}{1 - b + ab}$

$\rm \: = \: \: \dfrac{ 1 - b + ab}{1 - b + ab}$

$\rm \: = \: \: 1$

Hence, Proved

Concept Used :-

The Concept Used to solve this type of problem is to reduce the three variables to two variables.