Clarifying a question on polynomials Below is the question:
Determine the polynomials $f \in \mathbb R[X]$ for which there are $a,b \in \mathbb R, a \ne b$ , so that:

$(X-a)*f(X-b)=(X-b)*f(X-a)$.

When they say "for which there are $a,b \in \mathbb R$" , can it be deducted that $a$ and $b$ are fixed (for any $X$ they remain constant) ? 
To me they are not fixed because this is how I interpret it: polynomial f has a formula so that when being substituted in the above expression it result that a and b are real but NOT necessarily some particular values. The only difference between X and a,b would be that X can take any type of value, but a,b can only have real values.
I think it's a little vague my interpretation, so this is why I'd like to see yours. Thanks in advance!
 A: The question is asking to find all polynomials $f$ for which you can find $a,b\in\mathbb R$ with $a\ne b$ such that the displayed identity holds. The concrete numbers $a,b$ may very well depend on $f$. A priori, it may be that for some $f$ there is only one pair for which the identity holds, it may be that for some $f$ there are many such pairs, and it may also be that a pair that works for some $f$ does not work for some other $\hat f$. 

Let me address the question itself: Suppose that $f\in\mathbb R[x]$ is such that for some $a\ne b\in\mathbb R$ we have that $(x-a)f(x-b)=(x-b)f(x-a)$. Evaluating at $x=a$ makes the left-hand side $0$, so the right-hand side must vanish as well, and since $a\ne b$, then $x-a$ must be a factor of $f(x-a)$, that is, for some polynomial $g\in\mathbb R[x]$, we have
 $$ f(x)=xg(x). $$
Replacing in the given identity gives us that $(x-a)(x-b)g(x-b)=(x-b)(x-a)g(x-a)$. It follows that $g(x-b)=g(x-a)$. 
(Technical remark that can probably be skipped on a first reading: if it is not clear that the two polynomial expressions coincide, you could argue by thinking of the function that $g$ defines, and noticing that the equality holds when $x$ is evaluated as any number other than $a,b$. Since $g$ is a polynomial, it follows that the equality always holds, that is, the two expressions $g(x-b)$ and $g(x-a)$ coincide as polynomials. I am using that two distinct polynomials can only meet at finitely many points, which is a consequence of the fact that a nonzero polynomial has only as many roots as its degree.) 
It follows that $g$ is actually constant. Otherwise, if $r$ is a (possibly complex) zero of $g$, it follows from the identity that $r+(b-a)$ is also a zero (because $0=g(r)=g((r+b)-b)$; but $g(x-b)=g(x-a)$, so $g((r+b)-b)=g((r+b)-a)$). Iterating, we see that $r+2(b-a),r+3(b-a),\dots$ are all zeros of $g$. This is impossible, and we are forced to conclude that $g$ is constant, that is, for some $c\in\mathbb R$, we have that 
 $$ f(x)=cx. $$
Finally, note that for such an $f$ and any reals $a,b$ with $b\ne a$, we have that indeed $(x-a)f(x-b)=c(x-a)(x-b)=(x-b)f(x-a)$.
Curiously, this means that for any $f$ for which the identity holds, it holds for any pair $a,b$ whatsoever, which perhaps adds a bit to the initial confusion that prompted the question in the first place.

Two remarks, to conclude. First, the argument above is far from optimal but is perhaps as accessible as it gets with as few prerequisites as possible. More elegant approaches are certainly possible, and desirable.
Second, the fact that we are working over the reals was important for the argument above. More precisely, what I have actually shown is that polynomials of the form $cx$ are the only solutions to the given polynomial  equation $(x-a)f(x-b)=(x-b)f(x-a)$, $a\ne b$, when working in characteristic $0$. 
Other solutions are possible if we remove this last restriction. For instance, working over the field $\mathbb F_2$ ("that is", using arithmetic modulo $2$) we have that $f(x)=x^2(x+1)$ is a solution, with $a=0,b=1$.    
