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$ 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 $0$)

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    $\begingroup$ is $Q(x)=Ax+B$? $\endgroup$ – Dr. Sonnhard Graubner Aug 20 '17 at 13:11
  • $\begingroup$ are $x_1,x_2,x_3$ Zeros of $P(x)$? $\endgroup$ – Dr. Sonnhard Graubner Aug 20 '17 at 13:12
  • $\begingroup$ I think it is mx+n, the exercise doesn't say anywhere that the coefficients would be the same. @Dr.SonnhardGraubner $\endgroup$ – Alexander Aug 20 '17 at 13:13
  • $\begingroup$ yes, they are @Dr.SonnhardGraubner $\endgroup$ – Alexander Aug 20 '17 at 13:14
  • $\begingroup$ and $x_1,x_2,x_3$? what are These numbers? $\endgroup$ – Dr. Sonnhard Graubner Aug 20 '17 at 13:14
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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

$0$

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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, $$\sum_{i=1}^n\,\frac{Q\left(x_i\right)}{P'\left(x_i\right)}=q_{n-1}\,.$$

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