Fraction With Exponents!

Let $2015^a=x$ , $2015^b=y$ and $2015^c=z$ . Thus, $\sum \frac{1}{2015^{a-a}+2015^{a-b}+2015^{a-c}}=\sum \frac{1}{1+\frac{x}{y}+\frac{x}{z}}=\sum \frac{xy}{xy+yz+zx}=\frac{xy+yz+zx}{xy+yz+zx}=1$ Therefore the answer is $1$ .

Here $\sum$ is the cyclic sum .

Each denominator has a (different) common factor. Extracting these gives us the following expression: $\frac{2015^{-a}}{2015^{-a} + 2015^{-b} + 2015^{-c}} + \frac{2015^{-b}}{2015^{-a} + 2015^{-b} + 2015^{-c}} + \frac{2015^{-c}}{2015^{-a} + 2015^{-b} + 2015^{-c}}$

Now each denominator is the same, so this is simply: $\frac{2015^{-a} + 2015^{-b} + 2015^{-c}}{2015^{-a} + 2015^{-b} + 2015^{-c}} = 1$

Therefore, the answer is $\boxed{1}$ , regardless of the values of a, b, or c.

2015^a = x

2015^b = y

2015^c = z

1/(1 + x/y + x/z) + 1/(1 + y/x + y/z) + 1/(1 + z/y + z/x) =

yz/(yz + xz + xy) + xz/(yz + xz + xy) + yx/(yz + xz + xy) =(yz + xz + xy)/(yz + xz + xy)

= 1