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$f(x)=x^{n-1}+x^{n-2}+...+1$

I solved this issue and get $-1^{\frac{(n-1)(n-2)}{2}}*n^{n-2}$ as solution. I did it in a way of substitution n=1,2,3.. then i get my answer.

Now i want to solve it using dependence of resultant and discriminant, using next : $$D(f(x))=(-1)^{\frac{1}{2}n(n-1)}*\frac{1}{a_n}*R({f,f^{`}})$$ But it is look quite difficult to get $R({f,f^{`}})$. As a decision we can try to look at $g(x)=(x-1)*f(x)=x^n-1$. In this case much easier to calculate determinant of the matrix to calculate the $R(g,g^{`})$. But I have no idea how to bind $D(f(x))$ and $D(f(x)*(x-1))$

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In this case it may be simpler to express the discriminant in terms of the roots: en.wikipedia.org/wiki/Discriminant#Definition. –  lhf Jun 25 at 14:31
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$D(f(x)*(x-1))=D(f(x))*n$ –  Xoff Jun 25 at 14:31

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up vote 2 down vote accepted

As has been pointed out in the comments, there is a relation $D(f)=\prod_{i<j}(\alpha_i-\alpha_j)^2$ where the $\alpha_1,\ldots,\alpha_n$ are the roots of $f$. Then roots of $(x-1)f(x)$ are $1,\alpha_1,\ldots,\alpha_n$. Then the above relation gives $$D((x-1)f(x)) = (1-\alpha_1)^2\ldots (1-\alpha_n)^2 \prod_{i<j}(\alpha_i-\alpha_j)^.2 $$ But we can simplify this because $f(x) = (x-\alpha_1)\ldots (x-\alpha_n)$. So $(1-\alpha_1)^2\ldots (1-\alpha_n)^2 = f(1)^2$. Putting this together gives the relation $$D((x-1)f(x)) = f(1)^2 D(f(x)).$$

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