Let $x$ be an algebraic integer with minimal polynomial (over $\mathbb{Z}$) to be $F(X)=X^n+aX+b$. Then $$Disc(1,x,\cdots,x^{n-1})=(-1)^{n(n-1)/2}\mathbf{N}(F'(x)).$$ Aim: To compute $\mathbf{N}(F'(x))$:
(1) $F'(x)=nx^{n-1}+a$; call this element $y$. So $y=nx^{n-1}+a$.
(2) By equation of $x$, we get $y=-(n-1)a-nbx^{-1}$. Hence $x=-nb(y+(n-1)a)^{-1}$.
(3) The minimal polynomial of $y$ over $\mathbb{Z}$ is the numerator of $b^{-1}F(-nb(Y+(n-1)a)^{-1})$;[How?]
the result of the computation is $$(Y+(n-1)a)^n-na(Y+(n-1)a)^{n-1}+(-1)^nb^{n-1}.$$
Q. I did not get the assertion in (3); I did some computational verification, but my point is how it was directly asserted without doing explicit computations? In other words, there could be just one-two line clear justification of assertion in (3).
I did the computation as follows: $$x=\frac{-nb}{y+(n-1)a}.$$ Plug this $x$ in its equation $F(X)=0$ and clear the denominators, i.e. start expanding $$\Big{(}\frac{-nb}{y+(n-1)a}\Big{)}^n + a\Big{(}\frac{-nb}{y+(n-1)a}\Big{)}+b=0.$$
Ref. Algebraic theory of numbers, P. Samuel, p. 41.
Multiplying your equation in $y$ by $(y+(n-1)a)^n$ gives you an annihilating polynomial $(-nb)^n-anb(y+(n-1)a)^{n-1}+b(y+(n-1)a)^n$ of $y$ of degree $n$. But $\mathbf Q(x)=\mathbf Q(x^{-1})=\mathbf Q(y)$ by the definition of $y$, hence $y$ has degree $n$ and the above polynomial is actually a minimal polynomial of the algebraic number $y$. Multiplication by $b^{-1}$ gives you a monic annihilating polynomial of $y$ with coefficients in $\mathbf Z$, and you are done.