Question

show that this will equal to one
​

Answer 1

Concept used ---->

1) aᵐ⁻ⁿ = aᵐ a⁻ⁿ

2)

a⁻ᵐ = 1 /aᵐ

Solution----> LHS

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

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

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

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

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

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

$numerator \: and \: denomiator \: cancel \: out \: each \: other$

$= 1$

$= <strong><em>R</em></strong><strong><em>H</em></strong><strong><em>S</em></strong>$