algebra precalculus - Substitution to linear + nth power form

Given an arbitrary polynomial:

$$a_0 + a_1x + a_2x^2 ... a_nx^n$$

Does there exist a series of substitutions (or single substitution if you choose to combine them) that leaves this function in the form:

$$p_1w + p_2w^r$$

I am aware there are substitutions (referred to as polynomial depression) that leave the polynomial in the form:

$$p_1 + p_2w + rw^n$$

For example in this article:

http://en.wikipedia.org/wiki/Bring_radical#Bring.E2.80.93Jerrard_normal_form

Answer

Yes, given the general equation,

$$a_nx^n+a_{n-1}x^{n-1}+\dots+a_0 =0\tag{1}$$

you can use a deg $n-1$ Tschirnhausen transformation,

$$y =b_nx^{n-1}+b_{n-1}x^{n-2}+\dots b_1$$

to reduce (1) to binomial form,

$$y^n+c_0 = 0$$

Unfortunately, in general the unknowns $b_i$ entail solving an equation of at least $(n-1)!$ degree, hence is not in radicals for $n\geq5$. (For $n=4$, the system results in a solvable sextic.)

But you can eliminate, in radicals, the three terms $x^{n-1},x^{n-2},x^{n-3}$ simultaneously. A clear step-by-step description is given here.