Prove that: $\dfrac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\dfrac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\dfrac{1}{1+{{x}^{b-c}}+{{x}^{a-c}}}=1$ by using identity of exponents.

Hint: Take terms one by one and use the identity of exponents which is ${{x}^{m-n}}=\dfrac{{{x}^{m}}}{{{x}^{n}}}$ and simplify them and thus ass altogether to get desired results.

Complete step-by-step answer:

In the question we are given expression
$\dfrac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\dfrac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\dfrac{1}{1+{{x}^{b-c}}+{{x}^{a-c}}}$ and we have to prove that their value is 1.
So, let’s consider the expression which is
$\dfrac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\dfrac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\dfrac{1}{1+{{x}^{b-c}}+{{x}^{a-c}}}$
In this expression at first consider 1st term which is
$\dfrac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}$
In this we will use an identity of exponents which says that
$\dfrac{{{x}^{m}}}{{{x}^{n}}}={{x}^{m-n}}$
So we can write ${{x}^{b-a}}$ as $\dfrac{{{x}^{b}}}{{{x}^{a}}}$ and ${{x}^{c-a}}$ as $\dfrac{{{x}^{c}}}{{{x}^{a}}}$ so we can write 1st term as,
$\dfrac{1}{1+\dfrac{{{x}^{b}}}{{{x}^{a}}}+\dfrac{{{x}^{c}}}{{{x}^{a}}}}$
Now, we will take L.C.M. we get, \[\dfrac{1}{\dfrac{{{x}^{a}}+{{x}^{b}}+{{x}^{c}}}{{{x}^{a}}}}\] which is equal to \[\dfrac{{{x}^{a}}+{{x}^{b}}+{{x}^{c}}}{{{x}^{a}}}\].
Now, let’s consider 2nd term of the expression which is,
\[\dfrac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}\]
In this we will use an identity of exponents which say that,
\[\dfrac{{{x}^{m}}}{{{x}^{a}}}\ =\ {{x}^{m-a}}\]
So, we can write \[{{x}^{a-b}}\] as \[\dfrac{{{x}^{a}}}{{{x}^{b}}}\] and \[{{x}^{c-b}}\] as \[\dfrac{{{x}^{c}}}{{{x}^{b}}}\] so we can re-write 2nd term as, \[\dfrac{1}{1+\dfrac{{{x}^{a}}}{{{x}^{b}}}+\dfrac{{{x}^{c}}}{{{x}^{b}}}}\]
Now, we will take L.C.M. we get,
\[\dfrac{1}{\dfrac{{{x}^{b}}+{{x}^{a}}+{{x}^{c}}}{{{x}^{b}}}}\] which is equal to \[\dfrac{{{x}^{b}}}{{{x}^{b}}+{{x}^{a}}+{{x}^{c}}}\].
Now let’s consider 3rd term of the expression which is,
$\dfrac{1}{1+{{x}^{b-c}}+{{x}^{a-c}}}$
In this we will use an identity of exponents which say that,
$\dfrac{{{x}^{c}}}{{{x}^{a}}}={{x}^{m-n}}$
So we can write ${{x}^{b-c}}$ as $\dfrac{{{x}^{b}}}{{{x}^{c}}}$ and ${{x}^{a-c}}$ as $\dfrac{{{x}^{a}}}{{{x}^{c}}}$ so we can get third term as,
$\dfrac{1}{1+\dfrac{{{x}^{b}}}{{{x}^{c}}}+\dfrac{{{x}^{a}}}{{{x}^{c}}}}$
Now, we will take L.C.M we get,
$\dfrac{1}{\dfrac{{{x}^{c}}+{{x}^{b}}+{{x}^{a}}}{{{x}^{c}}}}$ which is equal to $\dfrac{{{x}^{c}}}{{{x}^{c}}+{{x}^{b}}+{{x}^{a}}}$
So now we can write the expression
$\dfrac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\dfrac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\dfrac{1}{1+{{x}^{b-c}}+{{x}^{a-c}}}$ as $\dfrac{{{x}^{a}}}{{{x}^{a}}+{{x}^{b}}+{{x}^{c}}}+\dfrac{{{x}^{b}}}{{{x}^{a}}+{{x}^{b}}+{{x}^{c}}}+\dfrac{{{x}^{c}}}{{{x}^{a}}+{{x}^{b}}+{{x}^{c}}}$
Which can be written as,
\[\dfrac{{{x}^{a}}+{{x}^{b}}+{{x}^{c}}}{{{x}^{a}}+{{x}^{b}}+{{x}^{c}}}\] which is equal to 1.
Hence, proved.

Note: We can simplify the expression by multiplying with ${{x}^{a}},{{x}^{b}},{{x}^{c}}$ to numerator and denominator to the respective terms and then simplify it to get the respective answer.