TITLE: (Variation of an old question) Are these functions polynomials?
QUESTION [5 upvotes]: This is a followup to the question here: How to show that the following function isn't a polynomial over Q?. 
As before, let $b_1,b_2,\dots$ be an enumeration of $\mathbb Q$. The question might be sensitive to the enumeration (but probably not).
Question 1: Suppose I define $$f_3(x) =  (x-b_1)^3 + (x-b_1)^3(x-b_2)^3 + \dots,$$
how do I show that this function is not a polynomial? I do not believe the answers to the old question extend to this case since they all seem to use the positivity of squares.
Can we show that $f(x)$ is not a polynomial?
There is a even more general version (and the one I was interested in from the beginning) that I was interested in, however I forgot to omit the trivial cases/there was some ambiguity in the second question, so I asked it again here: 
Is the following function a polynomial?

REPLY [4 votes]: $f_3(x)$ is not a polynomial function, which can be seen as follows. Call $b_n$ a champion if it exceeds all the earlier terms $b_1,\dots,b_{n-1}$. Clearly, there are infinitely many champions, and they tend to infinity. Let $m$ be fixed, and consider the champions $b_n$ with $n>m$. We get, as $b_n$ tends to infinity,
$$ f_3(b_n)>\prod_{k=1}^m (b_n-b_k)^3 = (1+o(1))\ b_n^{3m}. $$
This shows that $f_3$ has degree at least $3m$. As $m$ is arbitrary, the claim follows.