Question: Simplify: $\dfrac{1}{1+ a^{x-y} + a^{x-z}}+$$\dfrac{1}{1+ a^{y-x} + a^{y-z} }+$$\dfrac{1}{1+ a^{z-x} + a^{z-y}}$


Solution:
Given,

$= \dfrac{1}{1+ a^{x-y} + a^{x-z} }+\dfrac{1}{1+ a^{y-x} + a^{y-z}} +\dfrac{1}{1+ a^{z-x} + a^{z-y}}$

$\dfrac{1}{1+ \frac{a^x}{a^y}+ \frac{a^x}{a^z}} +\dfrac{1}{1+ \frac{a^y}{a^x }+ \frac{a^y}{a^z}} +\dfrac{1}{1+ \frac{a^z}{a^x}+ \frac{a^z}{a^y}}$

$= \dfrac{1}{\dfrac{a^y.a^z + a^x.a^z +a^x.a^y}{a^y.a^z}} + \dfrac{1}{\dfrac{a^x.a^z + a^y.a^z +a^y.a^x}{a^x.a^z}} + \dfrac{1}{\dfrac{a^x.a^y + a^y.a^z +a^x.a^z}{a^x.a^y}}$

$= \dfrac{a^y.a^z}{a^y.a^z + a^x.a^z +a^x.a^y} + \dfrac{a^x.a^z}{a^x.a^z + a^y.a^z +a^y.a^x} +\dfrac{a^x.a^y}{a^x.a^y + a^y.a^z +a^x.a^z}$

$= \dfrac{a^{y+z}}{a^{y+z}+ a^{x+z}+a^{x+y}} + \dfrac{a^{x+z}}{a^{x+z} + a^{y+z} +a^{y+x}} +\dfrac{a^{x+y}}{a^{x+y}+ a^{y+z} +a^{x+z}}$

[ Since, the denominators are common, we add the terms now. ]

$= \dfrac{a^{y+z} + a^{x+z} + a^{x+y}}{a^{x+y}+ a^{y+z} +a^{x+z}}$

$= \dfrac{(a^{x+y}+ a^{y+z} +a^{x+z})}{(a^{x+y}+ a^{y+z} +a^{x+z})}$

$= 1$
= Answer

Related Notes and Solutions:

Here is the website link to the notes of Indices.

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