Question

Prove that
$\frac{1}{1 + {x}^{a - b} } + \frac{1}{1 + {x}^{b - a} } = 1$
$\frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } + \frac{1}{1 + {x}^{b - c} + {x}^{a - c} } = 1$
​

Answer 1

ｓｏｌｕｔｉｏｎ:-

$\begin{gathered} \\ \scriptsize \begin{array}{:l:}\hline\\ \\ \\ \displaystyle\bf &\displaystyle\bf\left(\frac{1}{x^{a-b}}\right)^{\dfrac{1}{a-c}}\left(\frac{1}{x^{b-c}}\right)^{\dfrac{1}{b-a}}\left(\frac{1}{x^{c-a}}\right)^{\dfrac{1}{c-b}} \\\\ \\ \\ \displaystyle\bf \Rightarrow&\bf\bigg(x\bigg)^{\dfrac{1}{a-b} \times \dfrac{1}{a-c}}\bigg(x\bigg)^{\dfrac{1}{b-c} \times \dfrac{1}{b-a}}\bigg(x\bigg)^{\dfrac{1}{c-a} \times \dfrac{1}{c-b}} \\\\ \\ \\ \displaystyle\bf \Rightarrow&\bf\bigg(x\bigg)^{\dfrac{1}{a^{2}-a c-a b-b c}+\dfrac{1}{b^{2}-a b-b c+a c}+\dfrac{1}{c^{2}-b c-a c+a b}} \\ \\ \\ \\ \displaystyle\bf\Rightarrow&\bf\bigg( x \bigg) ^\dfrac{\left(a^{2}-a c-a b-c c-a^{2}+a c+a b+c\right)}{\left(a^{2}-a c-a b-b c\right)\left(b^{2}-a b-b c-a c\right)\left(c^{2}-b c-a c+a b\right)}\\\\\hline \end{array}\end{gathered}$

$\begin{gathered} \\ \scriptsize \begin{array}{:r:l:}\hline\\ \\ \\ \displaystyle\bf \Rightarrow&\bf\bigg(x\bigg)^{\dfrac{0}{\left(a^{2}-a c-a b-b c\right)\left(b^{2}-a b-b c-a c\right)\left(c^{2}-b c-a c+a b\right)}} \\ \\ \\ \\ \displaystyle\bf\Rightarrow&\bf\big(x\big)^{0} \\ \\ \\ \\ \displaystyle\bf\Rightarrow&\bf 1 \\\\ \\ \\ \displaystyle\bf \Rightarrow &\textbf { R.H.S }\\\\\hline \end{array} \end{gathered}$