While studying Ramanujan's Notebooks I stumbled upon the following entry :
$\frac{1}{(x^3-x)} = \frac{1}{2(x+1)} + \frac{1}{2(x-1)} - \frac{1}{x}$
Now, I can prove this identity while working on the RHS. However, if I were to prove it by expanding the LHS, how can I do it?
Since the denominator is $x(x^2-1)=x(x-1)(x+1)$ the partial fraction decomposition will be of the form: $ \frac{1}{x^3-x} = \frac{a}{x}+\frac{b}{x-1}+\frac{c}{x+1} \;\;\iff\;\; 1 = a(x-1)(x+1)+bx(x+1)+cx(x-1) \,$ where $a,b,c$ can be determined by identifying the coefficients, or for a shortcut:
$x=0 \implies 1=-a \implies a = -1\,$
$x=1 \implies 1 = 2b \implies b = \cfrac{1}{2}$
$x=-1 \implies 1 = 2c \implies c = \cfrac{1}{2}$
Very easily, combine the common denominators of the first two terms, and add to get $x/(x^2-1)$. Next, subtract $1/x$ from that and create a common denominator to yield the desired result!
We can prove it for RHS as you did by taking the common denominator of all of the fractions $$\frac{1}{2(x+1)} + \frac{1}{2(x-1)} - \frac{1}{x}$$ $$\frac{x^2-x}{2(x^3-x)} + \frac{x^2+x}{2(x^3-x)} - \frac{2x^2-2}{2(x^3-x)}$$ $$\frac{1}{x^3-x}$$ And there we have it. Proving it for the LHS can be done simply by taking the opposite steps. $$\frac{1}{x^3-x}$$ $$\frac{x^2-x+x^2+x-(2x^2-2)}{2(x^3-x)}$$ $$\frac{x^2-x}{2(x^3-x)} + \frac{x^2+x}{2(x^3-x)} - \frac{2x^2-2}{2(x^3-x)}$$ $$\frac{1}{2(x+1)} + \frac{1}{2(x-1)} - \frac{1}{x}$$ And there we have it again.
Generally if you can prove something by manipulating one side, you can simply reverse the steps to get to prove it by manipulating the other side. A good way of proving some equalities is actually to manipulate each expression until you get the same one. Just make sure not to divide by 0 when doing so!
