TITLE: Roots and relation between polynomials and their derivatives
QUESTION [10 upvotes]: This is probably easy but it might be interesting. Here goes $\dots$
Let $P\in\mathbb{R}[x]$ be a polynomial of degree $n>2$ and $P'=\frac{dP}{dx}$. If $x_1, x_2, \dots, x_n$ are the roots of $P(x)$, including multiplicities, consider the multi-variable expression
$$V_n(P)=\sum_{1\leq i<j\leq n}(x_i-x_j)^2.$$

Question: Is there a constant $c(n)$, independent of $P$, such that
  $V_{n-1}(P')=c(n)V_n(P)$? If so, what is $c(n)$?

REPLY [22 votes]: Suppose that we have 
$$P(x)=x^n-ax^{n-1}+bx^{n-2}+\cdots$$
where we can take $P$ to be monic since it doesn't affect $V_n(P)$. From Vieta's formula we have
$$a=\sum_{i=1}^n x_i \quad , \quad b=\sum_{1\le i<j\le n} x_ix_j$$
so we can find that $V_{n}(P)=(n-1)a^2-2nb$. Similarly we have 
$$V_{n-1}(P')=(n-2)\frac{(n-1)^2a^2}{n^2}-2(n-1)\frac{(n-2)b}{n}$$
$$=\frac{(n-1)(n-2)}{n^2}\left((n-1)a^2-2nb\right)=\frac{(n-1)(n-2)}{n^2}V_n(P)$$
so we get $c(n)=1-\frac{3}{n}+\frac{2}{n^2}$.