Question

(iv) a+b+c / (a-¹b-¹ + b-¹c-¹ + c-¹a-¹). Simplify it.

Answer is abc.


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Answer 1

Consider,

$\huge \bold{ \frac{a + b + c}{a {}^{ - 1} b {}^{ - 1} + b {}^{ - 1} {c}^{ - 1} {a}^{ - 1} } }$

$\bold{ = \frac{a + b + c}{ \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} } - - - - since \: x {}^{ - 1} = \frac{1}{x} }$

$\bold{ = \frac{a + b + c}{ \frac{c}{abc} + \frac{a}{abc} + \frac{b}{cab} }} \: \bold{ multiply \: and}$

divide each terms in the denominator by the missing terms. i.e., c,a,b respectively

$\huge \bold{ = \frac{ \frac{a + b + c}{c + a + b} }{abc} }$

$\bold{ = \frac{ \frac{1}{1} }{abc} }$

= abc

Hence proved.