Question

If x, y, z are positive real numbers and a, b, c are rational numbers, then the value of $\dfrac{1}{{1 + {x^{b - a}} + {x^{c - a}}}} + \dfrac{1}{{1 + {x^{a - b}} + {x^{c - b}}}} + \dfrac{1}{{1 + {x^{b - c}} + {x^{a - c}}}}$ is
A) -1
B) 0
C) 1
D) None of the above

Answer 1

Hint:
Firstly, simplify the terms in the form of ${x^{p - q}}$ as $\dfrac{{{x^p}}}{{{x^q}}}$ , where p, q = a, b, c as required.
Then, take the LCM in the denominators and simplify it further.
Thus, we get the required answer.

Complete step by step solution:
It is given that, x, y, z are positive real numbers and a, b, c are rational numbers.
Now, we are asked to find the value of $\dfrac{1}{{1 + {x^{b - a}} + {x^{c - a}}}} + \dfrac{1}{{1 + {x^{a - b}} + {x^{c - b}}}} + \dfrac{1}{{1 + {x^{b - c}} + {x^{a - c}}}}$ .
We can write $\dfrac{1}{{1 + {x^{b - a}} + {x^{c - a}}}} + \dfrac{1}{{1 + {x^{a - b}} + {x^{c - b}}}} + \dfrac{1}{{1 + {x^{b - c}} + {x^{a - c}}}}$ as
 $\dfrac{1}{{1 + {x^{b - a}} + {x^{c - a}}}} + \dfrac{1}{{1 + {x^{a - b}} + {x^{c - b}}}} + \dfrac{1}{{1 + {x^{b - c}} + {x^{a - c}}}} = \dfrac{1}{{1 + \dfrac{{{x^b}}}{{{x^a}}} + \dfrac{{{x^c}}}{{{x^a}}}}} + \dfrac{1}{{1 + \dfrac{{{x^a}}}{{{x^b}}} + \dfrac{{{x^c}}}{{{x^b}}}}} + \dfrac{1}{{1 + \dfrac{{{x^b}}}{{{x^c}}} + \dfrac{{{x^a}}}{{{x^c}}}}}$
 $
   = \dfrac{1}{{\dfrac{{{x^a} + {x^b} + {x^c}}}{{{x^a}}}}} + \dfrac{1}{{\dfrac{{{x^a} + {x^b} + {x^c}}}{{{x^b}}}}} + \dfrac{1}{{\dfrac{{{x^a} + {x^b} + {x^c}}}{{{x^b}}}}} \\
   = \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}}} \\
   = 1 \\
 $

Thus, we get the value of $\dfrac{1}{{1 + {x^{b - a}} + {x^{c - a}}}} + \dfrac{1}{{1 + {x^{a - b}} + {x^{c - b}}}} + \dfrac{1}{{1 + {x^{b - c}} + {x^{a - c}}}}$ as 1.

Note:
Some properties of exponents and powers:
1) \[{a^{m + n}} = {a^m} \cdot {a^n}\]
2) \[{a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}\]
3) \[\dfrac{1}{{{a^n}}} = {a^{ - n}}\]
4) \[{a^n}{b^n} = {\left( {ab} \right)^n}\]
5) $\sqrt[n]{a} = {a^{\dfrac{1}{n}}}$