$\alpha _n ^n-1=0$
$\alpha _n=e^{2 \pi i/n}$
$f(x_1,x_2,x_3,\ldots,x_n)=(x_1+\alpha _n x_2+ \alpha _n ^2 x_3+\cdots+\alpha _n ^{n-1} x_n)^n$
I have read in Jim Brown's paper on page 5 that Lagrange showed
If n=3 then $f(x_1,x_2,x_3)$ Maximum can have 2 different results with all permutations of $(x_1,x_2,x_3)$
If n=4 then $f(x_1,x_2,x_3,x_4)$ Maximum can have 3 different results with all permutations of $(x_1,x_2,x_3,x_4)$
If n=5 then $f(x_1,x_2,x_3,x_4,x_5)$ Maximum can have 6 different results with all permutations of $(x_1,x_2,x_3,x_4,x_5)$
but no proof how he did that result.
According to the Paper, It was an important result for insolvability of quintic via radicals. Thus I searched the paper of Lagrange (Lagrange's 1771 paper reflections on the Algebraic theory of Equations ) in the internet but I could not find it.
Jım Brown's paper does not mention the general solution for n.
What is the general formula of how many different values can have $f(x_1,x_2,x_3,\ldots,x_n)$ with all permutations of $(x_1,x_2,x_3,\ldots,x_n)?$ Any idea to find the general formula for n?
or if it is not possible for all n , at least to show a way how easily to proof for n=3,n=4 and n=5 (I tried to do that n=3 is relatively easy but need a lot calculation in classic approach as binomial expansion). Could you please help me how to approach the problem without using group theory? I need a proof in algebraic way. And also welcome all links that shows how Lagrange proved for n=3,n=4 and n=5.
Note:I try to understand deeply how Abel and Ruffini showed the insolubility of quintic via radicals. The problem is also related to my other question that shown that f is not symmetric function n>2. and $f(x_1,x_2,x_3,\ldots,x_n)=f(x_n,x_1,x_2,\ldots,x_{n-1})=f(x_{n-1},x_n,x_1,\ldots,x_{n-2})=.....=f(x_2,x_3,x_4,\ldots,x_n,x_1)$
(totally $n$ permutation of f is equal each other) it means at least n values are the same in total $n!$ all permutations of $(x_1,x_2,x_3,\ldots,x_n)$ .
Thanks a lot for your answers and your advises.
$UPDATE:$ I completed the Proof for $n=3$ I would like to share my way for $n=3$
All permutations for $n=3$ are:
$1)$-->$f(x_1,x_2,x_3)=(x_1+\alpha _3 x_2+ \alpha _3 ^2 x_3)^3$
$2)$-->$f(x_3,x_1,x_2)=(x_3+\alpha _3 x_1+ \alpha _3 ^2 x_2)^3=\alpha _3 ^3(x_3+\alpha _3 x_1+ \alpha _3 ^2 x_2)^3=(\alpha _3 x_3+\alpha _3 ^2 x_1+ x_2)^3$
$3)$-->$f(x_2,x_3,x_1)=(x_2+\alpha _3 x_3+ \alpha _3 ^2 x_1)^3=\alpha _3 ^3(x_2+\alpha _3 x_3+ \alpha _3 ^2 x_1)^3=(\alpha _3 x_2+ \alpha _3 ^2 x_3+x_1)^3$
$4)$-->$f(x_1,x_3,x_2)=(x_1+\alpha _3 x_3+ \alpha _3 ^2 x_2)^3$
$5)$-->$f(x_2,x_1,x_3)=(x_2+\alpha _3 x_1+ \alpha _3 ^2 x_3)^3=\alpha _3 ^3(x_2+\alpha _3 x_1+ \alpha _3 ^2 x_3)^3=(\alpha _3x_2+\alpha _3 ^2 x_1+ x_3)^3$
$6)$-->$f(x_3,x_2,x_1)=(x_3+\alpha _3 x_2+ \alpha _3 ^2 x_1)^3=\alpha _3 ^3 (x_3+\alpha _3 x_2+ \alpha _3 ^2 x_1)^3=(x_3\alpha _3+\alpha _3 ^2 x_2+ x_1)^3$
It can be easily seen that
(Permutation 1 = Permutation 3) and (Permutation 2 = Permutation 3)
Thus (Permutation 1 = Permutation 2 =Permutation 3)
(Permutation 4 = Permutation 6) and (Permutation 5 = Permutation 6)
Thus (Permutation 4 = Permutation 5 =Permutation 6)
If so for n=3, the function can have 2 different result {Permutation 1= $f(x_1,x_2,x_3)$ , Permutation 4 = $f(x_1,x_3,x_2)$)
To test with inputs: $x_1=1, x_2=2 ,x_3=0$
we know very well that $1+\alpha _3+\alpha _3 ^2=0$
$\alpha _3=e^{2 \pi i/3}=-\frac{1}{2}+ i\frac{\sqrt{3}}{2}$
Permutation 1: $f(x_1,x_2,x_3)=f(1,2,0)=(1+2\alpha _3)^3=1+6\alpha _3+12\alpha _3 ^2+8=-3-6\alpha _3=-3i\sqrt{3}$
Permutation 4: $f(x_1,x_3,x_2)=f(1,0,2)=(1+2\alpha _3 ^2)^3=1+6\alpha _3^2+12\alpha _3 +8=3+6\alpha _3=3i\sqrt{3}$
As you see in the example above .Permutation 1 and Permutation 4 cannot be the same always.