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.
 A: Observe that the coefficients of 1/(x-a), 1/(x-b), 1/(x-c) seem very cyclic, if not symmetric. This strongly suggests to appeal to the theory of partial fractions. In fact, if in the coefficient of 1/(x-a), we replace the a with x, and do the similar thing for the rest, we will always get 
$$ \frac {x} {(x-a)(x-b)(x-c)} $$
Hence, this fraction is equal to your expression through partial fractions. (Yay to avoiding tedious expansions.)

If you are in algebra precalc, it is likely that you have done partial fractions. There is also another solution by the remainder factor theorem. 
A: $$
\frac{a}{(a-b)(a-c)(x-a)}=\frac{a((a-c)-(a-b))}{(b-c)(a-b)(a-c)(x-a)}=\frac{a}{(b-c)(a-b)(x-a)} - \frac{a}{(b-c)(a-c)(x-a)}
$$
Thus,
$$
\frac{a}{(a-b)(a-c)(x-a)} + \frac{b}{(b-a)(b-c)(x-b)} + \frac{c}{(c-a)(c-b)(x-c)}=\frac{1}{(b-c)(b-a)}(\frac{b}{x-b}-\frac{a}{x-a})+ \frac{1}{(c-a)(c-b)}(\frac{c}{x-c}-\frac{a}{x-a})
$$
$$
=\frac{x}{(b-c)(x-a)(x-b)}-\frac{x}{(b-c)(x-a)(x-c)}=\frac{x}{(x-a)(x-b)(x-c)}
$$
A: This answer is to show that this can be done by hand without any particular clever trick, but by being clearheaded and organised. With the original expression being symmetric, you should expect the appearance of symmetric expressions in the expansion, and often a fair amount of cancellation.
If we put everything over a common denominator we get $$\frac {-a(b-c)(x-b)(x-c)-b(c-a)(x-a)(x-c)-c(a-b)(x-a)(x-b)} {(a-b)(b-c)(c-a)(x-a)(x-b)(x-c)}$$
Now we notice that the coefficient of $x^2$ in the numerator is $-a(b-c)-b(c-a)-c(a-b)=0$ and the constant coefficient is $-abc(b-c)-abc(c-a)-abc(a-b)=0$.
The coefficient of $x$ is $$a(b-c)(b+c)+b(c-a)(c+a)+c(a-b)(a+b)$$$$=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2$$
One way to factorise such an expression is to note that it is zero if $a=b$, or $b=c$ or $c=a$ so must have a factor $d(a-b)(b-c)(c-a)$ where d is a constant obtained by equating the coefficient of $ab^2$, for example.
Another way (if you don't have a better idea) is to treat it as a quadratic in $c$ so that $$(b-a)c^2+(a^2-b^2)c+ab(b-a)=(b-a)\left(c^2-(a+b)c+ab\right)=(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)$$

A further note - the original expression was symmetric. The common denominator had a factor $(a-b)(b-c)(c-a)$ which is skew symmetric, and this can only happen if this same factor, which is the basic skew symmetric expression in three variables, also appears as a factor in the numerator (otherwise the whole expression will fail to be symmetric). The factor has to cancel - and this helps us to know what we are looking for, and to imagine short cuts.
