$ \frac{Q\left(x_1\right)}{P'\left(x_1\right)}+\frac{Q\left(x_2\right)}{P'\left(x_2\right)}+\frac{Q\left(x_3\right)}{P'\left(x_3\right)}=? $ $ P\left(x\right)=x^3+ax^2+bx+c $
$ x_1,x_2,x_3$ distinctive in pairs(and are the roots of $P$)
$Q(x)$ is a first degree polynomial
$ \frac{Q\left(x_1\right)}{P'\left(x_1\right)}+\frac{Q\left(x_2\right)}{P'\left(x_2\right)}+\frac{Q\left(x_3\right)}{P'\left(x_3\right)}=? $
How should I approach this? I tried the long polynomial division on the first term, but didn't seem to get anything useful out of it. (the answer is $0$)
 A: Consider the polynomial
$$f(x):=\sum_{i=1}^3\,Q\left(x_i\right)\frac{\left(x-x_{i-1}\right)\left(x-x_{i+1}\right)}{P'\left(x_i\right)}\,,$$
where the indices are consider modulo $3$.  Then, $f(x)$ is a polynomial in $x$ of degree at most $2$ such that $f\left(x_i\right)=Q\left(x_i\right)$ for every $i=1,2,3$ (i.e., $f(x)$ is the Lagrange polynomial interpolating $Q(x)$ at $x=x_1$, $x=x_2$, and $x=x_3$).  Therefore, $f(x)=Q(x)$.  That is, the coefficient of $x^2$ in $f(x)$ is $0$.  This means
$$\sum_{i=1}^3\,\frac{Q\left(x_i\right)}{P'\left(x_i\right)}=0\,.$$

More generally, suppose that $P(x)$ is a monic separable polynomial in $x$ over a field $K$ of degree $n\in\mathbb{Z}_{>0}$.  Write $x_1,x_2,\ldots,x_n$ for the roots of $P(x)$ in the algebraic closure of $K$.  Let $Q(x)\in K[x]$ be a polynomial of degree at most $n-1$; that is, $Q(x)=\sum\limits_{j=0}^{n-1}\,q_j\,x^j$ for some $q_0,q_1,\ldots,q_{n-1}\in K$.  Then,
$$\sum_{i=1}^n\,\frac{Q\left(x_i\right)}{P'\left(x_i\right)}=q_{n-1}\,.$$
A: Hints (this is an elementary argument, see @Batominovski for deeper reasoning):
Recall that (since $P(x)$ is monic), it can be factored as:
$$P(x)=(x-x_1)(x-x_2)(x-x_3)$$ 
and 
$$P'(x)=(x-x_1)(x-x_2)+(x-x_1)(x-x_3)+(x-x_2)(x-x_3).$$
Therefore, 
$$
P'(x_1)=(x_1-x_2)(x_1-x_3).
$$
Now, continue in this way and combine fractions.  If you write out the sum (assuming $Q(x)=mx+n$), you should get:

$$\displaystyle\frac{mx_1+n}{(x_1-x_2)(x_1-x_3)}+\frac{mx_2+n}{(x_2-x_1)(x_2-x_3)}+\frac{mx_3+n}{(x_3-x_2)(x_3-x_1)}.$$

If you simplify, you should get

$0$

