Question

solve this question don't answer if u don't know ​

Answer 1

To Prove :-

LHS = RHS .

Solution :-

Firstly we will solve the LHS of Question ie :-

$\sf{ \implies \: \frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } + \frac{1}{1 + {x}^{b - c} + {x}^{a - c} } }$

Now here we have a formula that we are going to apply over here :-

${\boxed{\boxed{\sf{\implies {a}^{m-n} = \frac{{a}^{m}}{{a}^{n}} }}}}$

$\sf{ \implies \: \frac{1}{1 + \frac{ {x}^{b} }{ {x}^{a} } \: + \frac{ {x}^{c} }{ {x}^{a} } } + \frac{1}{1 + \frac{ {x}^{a} }{ {x}^{b} } + \frac{ {x}^{c} }{ {x}^{b} } } + \frac{1}{1 + \frac{ {x}^{b} }{ {x}^{c} } + \frac{ {x}^{a} }{ {x}^{c} } } }$

Now taking LCM of the digits which are in the denominator of 1 .

$\sf{ \implies \: \frac{1}{ \frac{ {x}^{a} + {x}^{b} + {x}^{c} }{ {x}^{a} } } + \frac{1}{ \frac{ {x}^{b} + {x}^{a} + {x}^{c} }{ {x}^{b} } } + \frac{1}{ \frac{ {x}^{c} + {x}^{b} + {x}^{a} }{ {x}^{c} } } }$

Now have a look over one more thing :-

$\sf{ \implies \: \frac{a}{ \frac{x}{y} } = \frac{a.y}{x} }$

Similarly using this concept in our solution also we'll get :-

$\sf{ \implies \frac{1. {x}^{a} }{ {x}^{a} + {x}^{b} + {x}^{c} } + \frac{1. {x}^{b} }{ {x}^{a} + {x}^{b} + {x}^{c} } + \frac{1. {x}^{c} }{ {x}^{a} + {x}^{b} + {x}^{c} } }$

Now again taking LCM of this above mentioned fraction :-

$\sf{ \implies \: \frac{ {x}^{a} + {x}^{b} + {x}^{c} }{ {x}^{a} + {x}^{b} + {x}^{c} } = 1 }$

Our RHS is 1 and also LHS became 1 .

Hence proved.

Answer 2

❏ SolutioN :

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$\blacksquare\:\:\footnotesize{\underline{\underline{To\: Prove}}}$

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

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L.H.S.

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

$\footnotesize{\text{Now from the index law we know that}\:\:x^{m-n}=\dfrac{x^m}{x^n}}$

$=\footnotesize{\dfrac{1}{1+\dfrac{x^b}{x^a}+\dfrac{x^c}{x^a}}+\dfrac{1}{1+\dfrac{x^a}{x^b}+\dfrac{x^c}{x^b}}+\dfrac{1}{1+\dfrac{x^b}{x^c}+\dfrac{x^a}{x^c}}}$

$=\footnotesize{\dfrac{1}{\dfrac{x^a+x^b+x^c}{x^a}}+\dfrac{1}{\dfrac{x^a+x^b+x^c}{x^b}}+\dfrac{1}{\dfrac{x^a+x^b+x^c}{x^c}}}$

$=\footnotesize{\dfrac{x^a}{x^a+x^b+x^c}+\dfrac{x^b}{x^a+x^b+x^c}+\dfrac{x^c}{x^a+x^b+x^c}}$

$=\footnotesize{\dfrac{\cancel{x^a+x^b+x^c}}{\cancel{x^a+x^b+x^c}}}$

$=\footnotesize{=1=R. H. S. \:(proved)}$

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