Question: If abc + 1 = 0, prove that: $\dfrac{1}{1-a-b^{-1}} +$$\dfrac{1}{1-b-c^{-1}} + $$\dfrac{1}{1-c-a^{-1}} = 1$

Solution:
Given,

$abc + 1= 0$
$or, abc = -1$

LHS
$= \dfrac{1}{1-a-b^{-1}} + \dfrac{1}{1-b-c^{-1}} + \dfrac{1}{1-c-a^{-1}}$

$= \dfrac{1}{1 - a - \dfrac{1}{b}} + \dfrac{1}{1-b-\dfrac{1}{c}} + \dfrac{1}{1-c - \dfrac{1}{a}}$

$= \dfrac{1}{\dfrac{b - ab - 1}{b}} + \dfrac{1}{\dfrac{c-bc-1}{c}} + \dfrac{1}{\dfrac{a-ac-1}}$

$= \dfrac{b}{b-ab-1} + \dfrac{c}{c-bc-1} + \dfrac{a}{a-ac-1}$

[ Multiply first term by 'c/c' ]

$= \dfrac{bc}{bc - abc -c} + \dfrac{c}{c-bc-1} + \dfrac{a}{a-ac-1}$

$= \dfrac{bc}{bc -(-1) -c} + \dfrac{c}{c-bc-1} + \dfrac{a}{a-ac-1}$

$= \dfrac{bc}{1 + bc -c} - \dfrac{c}{1+bc -c} + \dfrac{a}{a-ac-1}$

$= \dfrac{bc -c}{1+bc-c} + \dfrac{a}{a-ac-1}$

[ Multiply first term by 'a/a' ]

$= \dfrac{abc - ac}{a +abc - ac} +\dfrac{a}{a-ac-1}$

$= \dfrac{(-1) - ac}{a +(-1) -ac} +\dfrac{a}{a-ac-1}$

$= \dfrac{-1-ac}{a-ac-1} + \dfrac{a}{a-ac-1}$

$= \dfrac{-1-ac+a}{a-ac-1}$

$= \dfrac{(a-ac-1)}{(a-ac-1)}$

$= 1$
RHS


Related Notes and Solutions:

Here is the website link to the notes of Indices.

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