I am new here and of course I would like to be specific. I've found a general solution to the following system:
If $a$, $b$ and $c$ are given such that: $$ x+y+z=a $$ $$ x²+y²+z²=b $$ $$ xⁿ+yⁿ+zⁿ=c $$
Where $n=8$.
$x$, $y$ and $z$ could be devised using the method I have derived.
I just want to basically ask if this kind of thing is publishable. Thanks for your reply in anticipation.
I'm doubtful that this would be publishable unless the method is particularly neat or generalizable; it's not hard to find a method using matrices. Explicitly:
If a 3 by 3 matrix $A$ has eigenvalues $x, y, z$, then $A^8$ has eigenvalues $x^8, y^8, z^8$. Therefore, $tr(A^8) = x^8 + y^8 + z^8$. We can choose $A$ to be the companion matrix for a polynomial $x^3 - l x^2 - mx - n$. Then
$$x + y + z = tr(A) = l$$
$$x^2 + y^2 + z^2 = tr(A^2) = l^2 + 2m$$
$$x^8 + y^8 + z^8 = tr(A^8) = l^8 + 8 l^6 m + 20 l^4 m^2 + 16 l^2 m^3 + 2 m^4 + 8 l^5 n + 32 l^3 m n + 24 l m^2 n + 12 l^2 n^2 + 8 m n^2$$
So we set $l = a, m = \frac{b - a^2}{2}$, and solve the third equation to find $n$ (it's quadratic in $n$, so the solution won't be too difficult). Then $x, y, z$ are the eigenvalues of $A$, which can be determined by solving the above cubic.