Questions & Answers

Question

Answers

A. $ - 1$

B. $0$

C. $1$

D. None of these

Answer
Verified

Now we are going to use the law of exponents in expression $\left( 1 \right)$. that is, we are going to use the law ${x^{m - n}} = \dfrac{{{x^m}}}{{{x^n}}}$ in that expression. Therefore, we get

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

$ = \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}}}}}$

Let us simplify the above expression by taking LCM (least common multiple). Therefore, we get $\dfrac{1}{{\dfrac{{{x^a} + {x^b} + {x^c}}}{{{x^a}}}}} + \dfrac{1}{{\dfrac{{{x^b} + {x^a} + {x^c}}}{{{x^b}}}}} + \dfrac{1}{{\dfrac{{{x^c} + {x^b} + {x^a}}}{{{x^c}}}}}$

$ = \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}}}$

Let us rewrite the above expression by taking LCM. Therefore, we get

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

On cancellation of the term ${x^a} + {x^b} + {x^c}$, we get the required value and it is $1$.

Therefore, 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 equal to $1$.

Therefore, option C is correct.