Question

Solve this question​

Answer 1

Solution--->

1) Plzzz refer the attachment

2) In 2nd step we applied law of exponent as follows

aᵐ⁻ⁿ = aᵐ a⁻ⁿ

3) We take xᵃ , xᵇ , xᶜ common from denominator of first , second and third term .

4) Now we apply another law of exponent as follows

1 / xᵐ = x⁻ᵐ

5) Then we take LCM ( x⁻ᵃ + x⁻ᵇ + x⁻ᶜ ) .

6) In last step we cut numerator and denominator because they are same and get 1 , which is LHS

Answer 2

$\bold \red \star \: \large \underline{Solution:-}$

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

Talking LHS And try to Prove RHS:-

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

Taking Common :-

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

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

Can I write _______

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

So,

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