Let
$A = a+2000$, $B = b-2000$, $C = c+2000$, $D = d-2000$, $E = e+2000$, $F = f-2000$,
and
$$P(x) = (A + x)(B - x)(C + x) + (D - x)(E + x)(F - x).$$
$P(x)$ is a polynomial in $x$, and the left-hand side of the 3 equations given for the problem equals $P(-1)$, $P(0)$ and $P(1)$, respectively. The problem is asking to find the value $P(9)$.

Since we have
$$\begin{aligned} (A + x)(B - x)(C + x) &= -x^3 + \text{(terms with degree 2 or less)}, \\ (D - x)(E + x)(F - x) &= x^3 + \text{(terms with degree 2 or less)}, \end{aligned}$$
we can represent $P(x)$ in the form $P(x) = px^2 + qx + r$, with some real coefficients $p, q, r$. From the fact that we have $P(-1) = 1$, $P(0) = 10$ and $P(1) = 100$ we obtain
$$p - q + r = 1, \quad r = 10, \quad p + q + r = 100.$$
We can solve this simultaneous equations to get $(p, q, r) = (81/2, 99/2, 10)$, and conclude that
$$P(x) = \frac{81}{2}x^2 + \frac{99}{2}x + 10.$$
Substituting $9$ for $x$, we get
$$P(9) = \frac{81}{2} \cdot 81 + \frac{99}{2} \cdot 9 + 10 = 3736$$
for the answer.