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Find the value of the following expression $\dfrac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\dfrac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\dfrac{1}{1+{{x}^{b-c}}+{{x}^{a-c}}}$
(a) ${{x}^{a-b-c}}$
(b) $1$
(c) $0$
(d) $3$

Last updated date: 20th Jun 2024
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Hint: To evaluate this, we will first use the property of exponents to simplify denominator and then taking LCM (least common multiple) we will simplify the equation to obtain required results. Property of exponents that we will use are given as –
${{x}^{a+b}}={{x}^{a}}.{{x}^{b}}$ and ${{x}^{-a}}=\dfrac{1}{{{x}^{a}}}$, where $x$ is called the base variable and $a,b$ are powers in exponent.

Complete step by step answer:
We are given the equation as – $\dfrac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\dfrac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\dfrac{1}{1+{{x}^{b-c}}+{{x}^{a-c}}}$.
Now we will use property of exponents ${{x}^{a+b}}={{x}^{a}}.{{x}^{b}}$ to simplify denominator, we get –
As the equation still can used easily, so now let us use property of exponents ${{x}^{-a}}=\dfrac{1}{{{x}^{a}}}$ to simplify denominator, we get-
Let us take LCM of ${{x}^{a}}$ in denominator of first term, ${{x}^{b}}$ in denominator of second term, and LCM of ${{x}^{c}}$ in denominator of third term. Thus we will get –
Rearranging and simplifying, taking denominators of denominators to the numerator, we get –
Since denominators of all terms are same, so we can combine numerator to get –
Now, as we can see that both numerator and denominator are the same, so by dividing them we can get the answer as $1$.
Hence, solving the given equation, we find that $\dfrac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\dfrac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\dfrac{1}{1+{{x}^{b-c}}+{{x}^{a-c}}}=1$.

So, the correct answer is “Option B”.

Note: Students should do the calculations carefully step by step as calculations can be quite difficult. For solving property of exponents, ${{x}^{a+b}}={{x}^{a}}.{{x}^{b}}$ keep in mind that the base remains same. Also, LCM should be taken carefully in the denominator. Also, take care of the negative signs in the exponent, they change the meaning completely. Students should not get confused with ${{x}^{a-b}}$and${{x}^{b-a}}$, as both are different.