Given $f(x)$ and $g(x)$, find the positive difference between the respective sums of their roots.

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Here is another interesting math competition problem I found which I'd love to know how to do, especially given that I don't have much experience with products of sequences of terms. How would you solve this one?

Define $f(x)$ and $g(x)$ as follows:

$$f(x)=\prod_{i=0}^{2016}x^{i^{2017}}+\prod_{j=0}^{2017}(x^2+2016)^j$$

$$g(x)=\prod_{i=0}^{2016}x^{2017^i}+\left(\prod_{j=0}^{2017}x^{2j}\right)2016x$$

Find the positive difference between the respective sums of $f(x)$ and $g(x)$'s roots.

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Okay. First $f(x)$ and $g(x)$ are polynomials. Next, the sum of roots in a polynomial is given by the coefficient of the $x^{n-1}$th term. In $f(x)$ the highest power of $x$ is $\sum_{i=0}^{2016} i^{2017}$. Hence the sum of roots is the coefficient of $x$ with power $\sum_{i=0}^{2016} i^{2017}-1$. In $f(x)$ all the other powers of $x$ come from the second term. The highest power of $x$ in that is $\sum_{j=0}^{2017}2j$ i.e 2017x2018. This is definitely not even close to $\sum_{i=0}^{2016} i^{2017}-1$,the latter is bigger than $2016^{2017}$. Hence we conclude that there is no $x^{n-1}$ term which implies that the sum of roots in $f(x)$ is $0$. Applying the same argument to $g(x)$ we get the same result. Sum of roots in $g(x)$ is $0$. Hence the difference between the sum of their roots is $0$.