`1/(1+x^(b-a)+x^(c-a))+1/(1+x^(a-b)+x^(c-b))+1/(1+x^(b-c)+x^(a-c))=?`

Options

A. `x^(a-b-c)`
B. 1
C. 0
D. 3

Answer

To solve the expression

\[ \frac{1}{1+x^{b-a}+x^{c-a}} + \frac{1}{1+x^{a-b}+x^{c-b}} + \frac{1}{1+x^{b-c}+x^{a-c}}, \]

we will follow these steps:

### Step 1: Rewrite Each Fraction
We can rewrite each term in the expression using the property of exponents.

The first term can be rewritten as:
\[ \frac{1}{1 + x^{b-a} + x^{c-a}} = \frac{1}{1 + \frac{x^b}{x^a} + \frac{x^c}{x^a}} = \frac{1}{1 + \frac{x^b + x^c}{x^a}}. \]

Similarly, we can rewrite the second and third terms:
\[ \frac{1}{1 + x^{a-b} + x^{c-b}} = \frac{1}{1 + \frac{x^a}{x^b} + \frac{x^c}{x^b}} = \frac{1}{1 + \frac{x^a + x^c}{x^b}}, \]
\[ \frac{1}{1 + x^{b-c} + x^{a-c}} = \frac{1}{1 + \frac{x^b}{x^c} + \frac{x^a}{x^c}} = \frac{1}{1 + \frac{x^b + x^a}{x^c}}. \]

### Step 2: Find a Common Denominator
Next, we need to find a common denominator for the three fractions. The common denominator will be:
\[ (1 + x^{b-a} + x^{c-a})(1 + x^{a-b} + x^{c-b})(1 + x^{b-c} + x^{a-c}). \]

### Step 3: Combine the Fractions
Now we can combine the fractions over the common denominator:
\[ \frac{(1 + x^{a-b} + x^{c-b})(1 + x^{b-c} + x^{a-c}) + (1 + x^{b-a} + x^{c-a})(1 + x^{b-c} + x^{a-c}) + (1 + x^{b-a} + x^{c-a})(1 + x^{a-b} + x^{c-b})}{(1 + x^{b-a} + x^{c-a})(1 + x^{a-b} + x^{c-b})(1 + x^{b-c} + x^{a-c})}. \]

### Step 4: Simplify the Numerator
The numerator will simplify as we expand and combine like terms. Each of the terms will contribute to the overall sum, and after simplification, we will notice that the numerator will equal the denominator.

### Step 5: Final Result
After simplification, we find that the entire expression simplifies to:
\[ 1. \]

### Conclusion
Thus, the final answer is:
\[ \frac{1}{1+x^{b-a}+x^{c-a}} + \frac{1}{1+x^{a-b}+x^{c-b}} + \frac{1}{1+x^{b-c}+x^{a-c}} = 1. \]