Knowing $\prod_{i=1}^n (c_1 - a_i) = d_1$ can we determine $\prod_{i=1}^n (c_2 - a_i)$? For known set of real values $\{a_i\}$ and $c_1$, if we know
$$\prod_{i=1}^n (c_1 - a_i) = d_1$$
is there a way to determine
$$\prod_{i=1}^n (c_2 - a_i)$$
without having to redo the multiplications?
(Forgot to include additional relevant assumptions): 
All $a_i$ are strictly positive, and no $a_i$ is equal to $c_1$ or $c_2$; in fact both ${c_1,c_2}$ are greater than the $\max \{a_i\}$. The product is strictly positive. Thank you.
 A: Here is a way to rephrase your question: given a monic polynomial $p$ of degree $n$ with roots $a_1,\ldots,a_n$, and such that $p(c_1)=d_1$, find $p(c_2)$.
Of course this can always be done, as you can recover the coefficients for $p$ from the symmetric polynomials. Namely, you have 
$$
p(x)=\sum_{k=0}^n (-1)^{n}e_n(a_1,\ldots,a_n)\,x^{n-k}.
$$
Other than that, I don't think you can expect any kind of general answer. Consider for instance the case where $a_1=\ldots=a_n=0$. Then you are asking something like

If I now that $\pi^n=d_1$, find $e^n$. 

Which can be answer, but not in a "polynomial" way. 
A: Unless $n=1$, this is not possible in general. Let $P(x) = (x-a_1)(x-a_2)...(x-a_n)$, then you are asking is if there easy way to find $P(c_2)$, knowing only $P(c_1).$ Normally, you would compute the coefficients of $P(x)$ and then plug $c_2$ in to find $P(c_2).$ This is much worse than calculating $\prod_{i=1}^n(c_2-a_i)$ by $n-1$ multiplication. So I'd say multiplying out is actually the easiest way to compute $P(c_2),$ since you say you know the roots $a_i.$
It seems like you want to do "better" than performing $n-1$ multiplication and I highly doubt this can be done.  
