2016 is coming!

Note that all the coefficients including the constant are perfect cubes.

Let $P(x)=\sum_{k=0}^{2016}(2017-k)^3x^k$

Then the polynomial with roots $1-a_k$ for $1\le k\le 2016$ is

$Q(x)=P(1-x)=\sum_{k=0}^{2016}(2017-k)^3(1-x)^k$

But now we see the constant term of $Q(x)$ is clearly

$Q(0)=\sum_{k=0}^{2016}(2017-k)^3=\sum_{k=1}^{2017}k^3=(2035153)^2 mod1000 = 409$

@Otto Bretscher , how is the solution ?

Let $y = 1 - x \rightarrow x = 1-y$
Substituting in the given equation,
$f(y) = \sum_{r=1}^{r=2017} r^{3}(1-y)^{2017-r}$
The constant term in the binomial expansion of the $r^{th}$ term is $r^{3}$
Thus, the constant term(C) is given by C = $\sum_{r=1}^{2017}r^3$
Now, $\sum_{r=1}^{n}r^3 = \left(\dfrac{(n)(n+1)}{2}\right)^2$
Putting $n = 2017$
C = $\left(\dfrac{(2017)(2018}{2}\right)^2 = \left( 2017 \times 1009 \right)^2$
$C \equiv \left( 2017 \times 1009 \right)^2 \pmod{1000}$
$C \equiv \left( 17 \times9 \right)^2 \pmod{1000}$
$C \equiv 153^{2} \pmod{1000}$
$\therefore C \equiv 23409 \pmod{1000}$
Thus, the last 3 digits of C are 409.

@Dev Sharma you must mention that the new polynomial is a monic polynomial.