Let $P:\mathbb{R} \rightarrow \mathbb{R} $ , $ P(x)=x^3-mx^2-nx-p $
$ m,n,p \in \mathbb{R} $ and $a,b,c$ the roots of $P(x)$
Calculate $ S= \frac{(b+c)bc}{P'(a)} + \frac{(c+a)ca}{P'(b)} + \frac{(a+b)ab}{P'(c)} $
Using Viete's formulas we get :
$ a + b + c $ = m
$ ab+bc+ac $ = -n
$ abc $ =p
$ b+c $ = $m-a$ , $ c+a $ = $m-b$ , $ a+b $ = $m-c$
Knowing that $a,b,c$ are the roots of $P$ $\rightarrow $ $ P(x) = (x-a)(x-b) (x-c) $
$ P'(x)=(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b) $
$ P'(a)=(a-b)(a-c) $
$ P'(b)=(b-a)(b-c) $
$ P'(c)=(c-a)(c-b) $
After doing these I tried to rewrite S to an easier form but I think what I've done doesn't work at all.
There are 3 answers for $ S $ :
$ a ) $ $ S = m $
$ b ) $ $ S = m+n $
$ c ) $ $ S = m+n+p $
We will assume $a,b,c$ are distinct. Otherwise, at least one of $P'(a), P'(b), P'(c)$ vanishes and the expression at hand become ill defined.
When $a,b,c$ are distinct, we have following partial fraction decomposition:
$$\frac{1}{P(x)} = \sum_{\lambda \in \Lambda}\frac{1}{P'(\lambda)(x-\lambda)} \quad\text{ where }\quad \Lambda = \{ a, b, c \}\tag{*1}$$
Since $a + b + c = m$ and $abc = p$, we can rewrite the expression as $$S = \sum_{\lambda \in \Lambda} \frac{(m-\lambda)p}{\lambda P'(\lambda)} = mp \sum_{\lambda \in \Lambda}\frac{1}{\lambda P'(\lambda)}\; -\; p\sum_{\lambda\in \Lambda} \frac{1}{P'(\lambda)} $$
Take $x = 0$ in $(*1)$, we obtain
$$\sum_{\lambda \in \Lambda} \frac{1}{\lambda P'(\lambda)} = - \frac{1}{P(0)} = \frac{1}{p}$$ Multiply $(*1)$ by $x$ and send $x$ to $\infty$, we obtain
$$\sum_{\lambda \in \Lambda}\frac{1}{P'(\lambda)} = \lim_{x\to\infty} \sum_{\lambda\in\Lambda}\frac{x}{P'(\lambda)(x-\lambda)} = \lim_{x\to\infty} \frac{x}{P(x)} = 0$$
Combine these, we can deduce $$S = mp \cdot \frac{1}{p} - p \cdot 0 = m = a+b+c$$