Answer

Verified

404.7k+ views

**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 –

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

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-

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

Let us 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 –

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

Rearranging and simplifying, taking denominators of denominators to the numerator, we get –

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

Since denominators of all terms are same, so we can combine numerator to get –

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

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.

Recently Updated Pages

Differentiate between Shortterm and Longterm adapt class 1 biology CBSE

How do you find slope point slope slope intercept standard class 12 maths CBSE

How do you find B1 We know that B2B+2I3 class 12 maths CBSE

How do you integrate int dfracxsqrt x2 + 9 dx class 12 maths CBSE

How do you integrate int left dfracx2 1x + 1 right class 12 maths CBSE

How do you find the critical points of yx2sin x on class 12 maths CBSE

Trending doubts

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Name 10 Living and Non living things class 9 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

List some examples of Rabi and Kharif crops class 8 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE