Simplifying $\frac{a}{(a-b)(a-c)(x-a)}+\frac{b}{(b-c)(b-a)(x-b)}+\frac{c}{(c-a)(c-b)(x-c)}$ We need to simplify $$\dfrac{a}{(a-b)(a-c)(x-a)}+\dfrac{b}{(b-c)(b-a)(x-b)}+\dfrac{c}{(c-a)(c-b)(x-c)}$$
The biggest problem is that the above expression has four variables.I transformed the expression into  $$\dfrac{a}{-(a-b)(c-a)(x-a)}+\dfrac{b}{-(b-c)(a-b)(x-b)}+\dfrac{c}{-(c-a)(b-c)(x-c)}$$
 and then tried to add them up,but as you can probably guess,it yielded something almost impossible to handle.I tried to factorize it using standard techniques,but that was fruitless too.A hint will be appreciated.
N.B: The chapter deals with Remainder-factor theorem,partial fractions,factoring cyclic homogenous polynomials and manipulating algebraic expressions.

$$
\frac{a}{(a-b)(a-c)(x-a)}=\frac{a((a-c)-(a-b))}{(b-c)(a-b)(a-c)(x-a)}=\frac{a}{(b-c)(a-b)(x-a)} - \frac{a}{(b-c)(a-c)(x-a)}
$$
Thus,
$$
\frac{a}{(a-b)(a-c)(x-a)} + \frac{b}{(b-a)(b-c)(x-b)} + \frac{c}{(c-a)(c-b)(x-c)}=\frac{1}{(b-c)(b-a)}(\frac{b}{x-b}-\frac{a}{x-a})+ \frac{1}{(c-a)(c-b)}(\frac{c}{x-c}-\frac{a}{x-a})
$$
$$
=\frac{x}{(b-c)(x-a)(x-b)}-\frac{x}{(b-c)(x-a)(x-c)}=\frac{x}{(x-a)(x-b)(x-c)}
$$
