Question

prove that:- 1/1+x^b-a +x^c-a + 1/ 1+x^a-b+x^c-b +1/1+x^b-c+x^a-c = 1​

Answer 1

SOLUTION :

$\sf \frac{1 }{1 + {x}^{(b - a)} + {x}^{(c - a)} } + \frac{1}{1 + {x}^{(a - b)} + {x}^{(c - b)} } + \frac{1}{1 + {x}^{(b - c)} {x}^{(a - c)} }$

$= \sf \frac{ {x}^{a} }{ {x}^{a} (1 + {x}^{(b - a)} + {x}^{(c - a)} )} + \frac{ {x}^{b} }{ {x}^{b} (1 + {x}^{(a - b)} + {x}^{(c - b)} ) } + \frac{ {x}^{c} }{ {x}^{c}( 1 + {x}^{(b - c)} {x}^{(a - c)} ) }$

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

$= \sf \: \frac{ \cancel{( {x}^{a} + {x}^{b} + {x}^{c} ) }}{ \cancel{( {x}^{a} + {x}^{b} + {x}^{c} })} =1$

[Proved]

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Answer 2

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$\: \bigstar \large \green { \bold{ \underline{ \underline{correct \: answer}}}}$

$\: \: \\ \: \: \ggg \bold{ \frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } } + \frac{1}{1 + {x}^{b - c} + {x}^{a - c}}$

$\: \: \bigstar \large \orange { \bold{ \underline{step \: by \: step \: explaination}}} :$

☑ LHS:

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

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

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

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

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

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

☑ Rhs =1

LHS =RHS, Hence proved

$\: \bigstar \displaystyle \bf{ \frac{1}{1 + {x}^{b - a} + {x}^{c - a} } } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } + \frac{1}{1 + {x}^{b - c} + {x}^{a - c} } = 1$

$\clubsuit \boxed{ \bf{ \frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } + \frac{1}{1 + {x}^{b - c} + {x}^{a - c} } = 1 }}$