Discriminant of a trinomial 
Let $b,c \in \mathbb{Z} $ and let $n \in \mathbb{N} $, $n \ge 2. $ Let $f(x) = x^{n} -bx+c$. Prove that $$\hbox{disc} (f(x)) = n^{n }c^{ n-1}-(n-1)^{n-1 }b^{n }.$$

Here $\hbox{disc} (f(x)) = \prod_{i} f'(\alpha_{i} )$ where $\alpha_{1}, \dots, \alpha_{n}$ are the roots of $f(x)$.
After some calculations I obtained $\hbox{disc} (f(x)) = \frac{\prod_{i} \alpha_{i}(n-1)b \ - \ nc }{\prod_{i} \alpha_{i}} $, but I'm afraid this is the wrong way.
 A: Here is a brute force approach:
$f'(x) = nx^{n-1} - b$, and we want to compute $\prod_i f'(\alpha_i)$. We do this by looking for the minimal polynomial with roots $\alpha_i^{n-1}$.
Note that
$$\begin{array}%
x^n - bx+c = 0 &\Leftrightarrow x(x^{n-1} - b) = -c \\
&\Leftrightarrow x^{n-1} (x^{n-1} - b)^{n-1} = (-1)^{n-1}c^{n-1}
\end{array}$$
let $y_i = \alpha_i^{n-1}$, and $z_i = f'(\alpha_i) = ny_i - b$. We have found the minimal polynomial for $y_i$:
$$y(y-b)^{n-1} = (-1)^{n-1}c^{n-1}$$
and we want the product $\prod_i z_i$. Consider the change of variable $z = ny-b$, i.e. $y = \frac{z+b}{n}$. Substitute into the above equation, we get
$$\frac{z+b}{n} \left(\frac{z - (n-1)b}{n}\right)^{n-1} = (-1)^{n-1}c^{n-1} \\
\Leftrightarrow (z+b)(z-(n-1)b)^{n-1} - (-1)^{n-1} n^n c^{n-1} = 0$$
The constant term of this polynomial in $z$ is $(-1)^n \prod_i z_i$, therefore
$$\prod_i z_i = (-1)^n \left((-1)^{n-1}b^n(n-1)^{n-1} - (-1)^{n-1} n^n c^{n-1}\right) = n^n c^{n-1} - (n-1)^{n-1}b^{n}$$
A: Two solutions. The first is based on resultants. As I wasn't 100% confident in my handling of the leading coefficients, I calculated the discriminant also using the definition. I am keeping the first (worse) solution for "educational purposes".

A promising way is to use the description of the discriminant as a resultant of $f(x)$ and $f'(x)$. The resultant is (when non-zero) also a generator of the ideal $I=(f(x),f'(x))\cap\mathbb{Z}$ and can be calculated with steps like the Euclidean algorithm, where we need to take care to keep everything as integers (necessitating appropriate scaling). 
Here we first calculate
$$
r_1(x)=nf(x)-xf'(x)= -b(n-1)x +cn
$$
and are "lucky" to get a linear remainder. Proceeding we see that
$$
g(x)=(cn)^{n-1}-[b(n-1)x]^{n-1}
$$
is a multiple of $r_1(x)$ by a polynomial in $\mathbb{Z}[x]$. We can then eliminate $x$ by calculating the combination
$$
r_2(x)=(b(n-1))^{n-1}f'(x)+ng(x)=n^nc^{n-1}-(n-1)^{n-1}b^n.
$$

Here's a more convincing calculation. Let us write
$$
\begin{aligned}
f(x)&=(x-\alpha_1)(x-\alpha_2)\cdots (x-\alpha_n),\\
f'(x)&=n(x-\beta_1)(x-\beta_2)\cdots (x-\beta_{n-1}).
\end{aligned}
$$
Here the numbers $\beta_j, j=1,2,\ldots,n-1,$ are the zeros of $f'(x)$, i.e. we have $\beta_j=\zeta^j K$, where $K=\root{n-1}\of {b/n}$ and $\zeta=e^{2\pi i/(n-1)}$ is an appropriate root of unity.
Now
$$
\begin{aligned}
\operatorname{disc}(f)&=\prod_i f'(\alpha_i)=n^n\prod_{i,j}(\alpha_i-\beta_j)\\
&=n^n(-1)^{n(n-1)}\prod_{i,j}(\beta_j-\alpha_i)\\
&=n^n\prod_jf(\beta_j).
\end{aligned}
$$
We can calculate
$$
f(\beta_j)=\beta_j(\beta_j^{n-1}-b)+c=c-(b-K^{n-1})\beta_j=c-\frac{b(n-1)K}n\zeta^j.
$$
so
$$
\operatorname{disc}(f)=n^n\prod_j(c-\frac{b(n-1)K}n\zeta^j).
$$
We can  use the factorization
$$
u^{n-1}-v^{n-1}=\prod_j(u-\zeta^jv)
$$
with $u=c$ and $v=b(n-1)K/n$ to evaluate this. The result is
$$
\operatorname{disc}(f)=n^n\left(c^{n-1}-\left(\frac{b(n-1)K}n\right)^{n-1}\right)
=n^n\left(c^{n-1}-\frac{(n-1)^{n-1}b^n}{n^n}\right),
$$
which equals the claimed formula.
