This is a followup to the question here: [How to show that the following function isn't a polynomial over Q?](https://mathoverflow.net/questions/288259/how-to-show-that-the-following-function-isnt-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?](https://mathoverflow.net/questions/288346/is-the-following-function-a-polynomial)



(Variation of an old question) Are these functions polynomials?
$f\_3(x)$ 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.


