$a$, $b$, $c$ and $x$ are literals. They constructed an expression to express a quantity in algebraic form and it is given that the algebraic form expression has to be simplified to obtain the required quantity.

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

In this problem, $x$ is a common base to six exponential terms and the exponent of each term is the subtraction of any two of three literals $a$, $b$ and $c$.

The exponent of each term in denominator of each term in this expression is in subtraction form. Use division rule of exponents to express each term in division form of two exponential terms having same base.

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

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

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

Arrange the three exponential terms in denominator in an order to simplify it easily.

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

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

$\require{cancel} = \dfrac{\cancel{x^a+x^b+x^c}}{\cancel{x^a+x^b+x^c}}$

The quantity in numerator and denominator is same. So, the quotient of them is one.

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

The problem can also be simplified in another method.

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

Observe the formation of each term in denominator of each term of this expression. The first term is one and the remaining two terms have a common base but each exponential term contains a quantity which can be obtained by the subtraction of two literals.

Try to write the number $1$ in exponential form similar to remaining to exponential terms by using zero exponent rule.

$(1)\,\,\,\,\,\,$ $1 = x^0 = x^{a-a}$

$(2)\,\,\,\,\,\,$ $1 = x^0 = x^{b-b}$

$(3)\,\,\,\,\,\,$ $1 = x^0 = x^{c-c}$

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

Apply the product rule of exponents and express each exponential term as a product of two exponential terms.

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

Take, $x^{-a}$, $x^{-b}$ and $x^{-c}$ common from each term of denominators of three terms of the expression.

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

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

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

$= \require{cancel} = \dfrac{\cancel{x^a+x^b+x^c}}{\cancel{x^a+x^b+x^c}}$

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

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