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 $c(n)$?
> 
> 
> 



Roots and relation between polynomials and their derivatives
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 $c(n)=1-\frac{3}{n}+\frac{2}{n^2}$.


