Fractions expansion is it possible to trasform this expression:
a,b,c constant and x a variable.
$$\frac{1}{(x-a)^2}+\frac{1}{(x-b)^2}+\frac{1}{(x-f)^2}$$
in a form that has the variable x outside the the parenthesis:
something like this:
$$(\cdots)x^4+ (\cdots)\frac{1}{x^4}+ (\cdots)x^3 + \cdots$$
I'm not looking for a series expantion, not an approximation, just to have somethin with all the constant coefficents that multiply or divide x.
Is not an exercise, is that I'm not very good in math and I cannot find a way to simplify/expand this function.
for example this (expanded form):
$$\tiny\frac{(a^2 b^2-2 a^2 b x+a^2 f^2-2 a^2 f x+2 a^2 x^2-2 a b^2 x+4 a b x^2-2 a f^2 x+4 a f x^2-4 a x^3+b^2 f^2-2 b^2 f x+2 b^2 x^2-2 b f^2 x+4 b f x^2-4 b x^3+2 f^2 x^2-4 f x^3+3 x^4)}{((a-x)^2 (x-b)^2 (x-f)^2)}$$
is not what I'm looking for.
What I'm asking first of all is: is possible? it can be that is not even possible.
What I want to do is pretty simple:
I want to simulate a vector field to simulate the behaviour of n particles.
The particles interacts with the others by this formula
$$\frac{1}{(\text{PositionParticle}_x-\text{PositionParticle}_{nx})^2}$$
$$\frac{1}{(\text{PositionParticle}_y-\text{PositionParticle}_{ny})^2}$$
which mean that the vector field function is:
$$\small f(x) = \frac{1}{(\text{PositionParticle}_x-\text{PositionParticle}_{n0})^2}+\frac{1}{(\text{PositionParticle}_x-\text{PositionParticle}_{n1})^2}+(\cdots)$$
same for $y$
computing this require a lot of calculation why the computer must access the array where there are stored the variable $n\times n$ times.
a faster way to do this would be to calculate the vector field having the coefficients summarized and stored and then passing them to modify the $x$ variable.
 A: One approach is to note that $$\frac{1}{(x-a)^2}=\sum_{n=0}^{\infty}a^{-2-n}(1+n)x^n=\frac{1}{a^2}+\frac{2x}{a^3}+\frac{3x^2}{a^4}\ldots$$
which converges for $|x| \lt a$.  You can add together one for each of your fractions.  Is that what you are looking for?
Added:  If $|x| \gt a$ you can also get an expression, but the $x$ will be in the denominator as it will be an expansion around $\infty$.  It looks similar:
$$\frac{1}{(x-a)^2}=\sum_{n=0}^{\infty}x^{-2-n}(1+n)a^n=\frac{1}{x^2}+\frac{2a}{x^3}+\frac{3a^2}{x^4}\ldots$$
A: It's impossible with rational functions, i.e. quotients of polynomials (with complex coefficients). Putting everything over a common denominator, we obtain the following equality of fractions, where we may assume that the LHS fraction is in lowest terms, hence $\rm\:n \ge 1\:$ is minimal
$$\rm \dfrac{f(x)}{x^n}\ =\ \frac{h(x)}{g(x)} $$
Above $\rm\:x\:$ does not divide $\rm\:f(x)\:$ since the fraction is in lowest terms, and also $\rm\:x\:$ does not divide $\rm\:g(x) = ((x-a)(x-b)(x-c))^2\:$ by $\rm\:a,b,c\ne 0\:.\:$ Equivalently $\rm\:f(0)\ne 0\ne g(0) = (abc)^2.\:$ Therefore, cross multiplying we obtain 
$$\rm f(x)\:g(x)\ =\ x^n\:h(x)$$
which when evaluated at $\rm\:x = 0\:$ yields $\rm\:f(0)\:g(0)= 0\:$ contra $\rm\:f(0)\ne 0\ne \:g(0)\:.$  Said equivalently, since the first fraction is in lowest terms, then its denominator $\rm\:x^n\:$ must divide the denominator of any equivalent fraction, thus $\rm\:x^n\:$ divides $\rm\:g(x)\:,\:$ so $\rm\:g(0) = abc = 0\:,\:$ so at least one of $\rm\:a,b,c\:$ is $\rm\:0\:.$ 
This is a special case of the unique fractionization property for polynomials over a field (which follows since they possess a Euclidean algorithm). This specific case is a bit simpler: it's equivalent to the fact that $\rm\:x\:$ is prime, i.e. $\rm\:x\:|\:f(x)\:g(x)\:$ implies $\rm\:x\:|\:f(x)\:$ or $\rm\:x\:|\:g(x)\:;\: $ or, said equivalently, that $\rm\:f(0)\:g(0) = 0\:$ implies $\rm\:f(0)=0\:$ or $\rm\:g(0)=0\:.\:$ Thus it is equivalent to the fact that the product of nonzero coefficients remains nonzero, i.e. that the coefficient ring is an integral domain. The algebraic structure at the heart of the above comes to the fore when one studies factorization theory in abstract algebra.
