How/why $\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}$ turn into (negative fraction) $-\frac{x+a}{x²a²}$?

Question

I have that the solution of $\left(\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}\right)$ is $-\frac{x+a}{x²a²}$, but when I try to solve it I get (positive) $\frac{x+a}{x²a²}$.

The way I've been getting it is by making the denominators equals $\left(\frac{1}{x²(x-a)}\frac{a²}{a²}\right)-\left(\frac{1}{a²(x-a)}\frac{x²}{x²}\right)$ which gives $\frac{a²-x²}{x²a²(x-a)}$ that by factoring the nominator is $\frac{(a-x)(a+x)}{x²a²(x-a)}$, allowing me to cancel $\frac{(a-x)}{(a-x)}$, finishing as $\frac{x+a}{x²a²}$.

Where it should've become negative?

OBS: I don't know if matters but it's all part of a limits problem that goes like $\lim_{x\to a}\left(\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}\right)=-\frac{x+a}{x²a²} = \frac{-2}{a³}$, but my problem is just where it became negative as it's suppose to become.

Answer 1

The correct cancellation is $$ \frac{a-x}{x-a}=-1 $$ since $a-x=(-1)(x-a)$.

Answer 2

$$\frac{1}{x^2(x - a)} - \frac{1}{a^2(x - a)}$$ $$\frac{1}{(x-a)}\left(\frac{1}{x^2} - \frac{1}{a^2}\right)$$ $$\frac{1}{(x-a)}\left(\frac{a^2 - x^2}{x^2a^2}\right)$$ $$\frac{1}{(x-a)}\left(\frac{(a + x)(a - x)}{x^2a^2}\right)$$ $$\frac{1}{(x-a)}\left(\frac{(a + x) -1(x - a)}{x^2a^2}\right)$$ $$\frac{-(a + x)}{x^2a^2}$$