How to transform a general higher degree five or higher equation to normal form? This is a continuation of How to solve fifth-degree equations by elliptic functions?
How to transform a general higher degree five or higher equation to normal form?  For example,  a quintic equation to Bring-Jerrard form?
 A: To reduce the general quintic,
$$x^5+ax^4+bx^3+cx^2+dx+e=0\tag{1}$$
to Bring-Jerrard form,
$$x^5+x+k = 0\tag{2}$$
is done in two steps. 
Step 1: Transform $(1)$ to principal quintic form (which is missing the $x^4,x^3$ terms) using a quadratic Tschirnhausen transformation,
$$y=x^2+mx+n\tag{3}$$
and eliminate $x$ between $(1)$ and $(3)$ using resultants. Nowadays, this is easily done by Mathematica or Maple. In wolframalpha.com, the command is,
  Collect[Resultant[x^5+ax^4+bx^3+cx^2+dx+e, y-(x^2+mx+n), x],y] 

which eliminates $x$ and collects the new variable $y$ yielding,
$$y^5+c_1y^4+c_2y^3+c_3y^2+c_4y+c_5=0\tag{4}$$
where,
$$c_1 = -a^2 + 2 b + a m - 5 n$$
$$c_2 = b^2 - 2 a c + 2 d - a b m + 3 c m + b m^2 + 4 a^2 n - 8 b n - 4 a m n + 10 n^2$$
and so on. The two unknowns $m,n$ allow you to eliminate two $c_i$. One can see that solving $c_1 = c_2 = 0$ will need only a quadratic. Thus, $(1)$ becomes the principal quintic form,
$$y^5+uy^2+vy+w=0\tag{5}$$
Step 2: To transform this to Bring-Jerrard, the impulse is to use a cubic Tschirnhausen. But this involves a composition of 1st, 2nd, 3rd-deg equations which will result in a sextic. Bring and Jerrard cleverly found a way around that by using a quartic Tschirnhausen, 
$$z = y^4+py^3+qy^2+ry+s\tag{6}$$
and the extra parameter prevents elevation of degree. Eliminating $y$ between $(5)$ and $(6)$, we get,
$$z^5+d_1z^4+d_2z^3+d_3z^2+d_4z+d_5=0\tag{7}$$
where,
$$d_1 = -5 s + 3 p u + 4 v$$
$$d_2 = 10 s^2 - 12 p s u + 3 p^2 u^2 - 3 q u^2 + 2 q^2 v - 16 s v + 5 p u v + 
  6 v^2 + 5 p q w - 4 u w + r \color{brown}{(3 q u + 4 p v + 5 w)}$$
and so on. Similar to the first step, solving $d_1 = d_2 = 0$ will need only a quadratic. One then uses 3 variables $p,q,s$ to solve the 3 equations,
$$\color{brown}{3 q u + 4 p v + 5 w} = 0\tag{8}$$
$$d_1 = d_2 = 0\tag{9}$$ 
But notice that by solving $(8)$, it causes $r$ to disappear from $d_2$ and it remains a free parameter. Since the third term of $(7)$ has form,
$$d_3 = e_3r^3+e_2r^2+e_1r+e_0$$
where the $e_i$ are polynomials in the other variables, one can then use $r$ to solve $d_3 =0$ merely as a cubic. (If the general quintic wasn't reduced to principal form first, it would be harder to make $r$ disappear from $d_2$. Bring and Jerrard were clever, weren't they?)
What remains is,
$$z^5+d_4z+d_5 = 0$$
We can make a further simplification $d_4 = \pm1$ by scaling variables $z = t/f$,
$$t^5+d_4f^4t+d_5f^5 = 0$$
and solving for $f$ in $d_4f^4 =\pm1$. Thus, we end up with the Bring-Jerrard quintic,
$$t^5\pm t+k = 0\tag{10}$$
P.S. The same approach can be used to eliminate the $x^{n-1},x^{n-2},x^{n-3}$ terms simultaneously from the general equation of degree $n>3$.
