Find the GCD of $10$ integers 
Find the greatest common divisor of the following ten integers: 
  \begin{align}
& 2000^3+3\cdot2000^2+2\cdot2000,\  2001^3+3\cdot2001^2+2\cdot2001,\  \dots, \\[5pt]
& \dots, \  2008^3+3\cdot2008^2+2\cdot2008,\  2009^3+3\cdot2009^2+2\cdot2009.
\end{align}

How are these numbers related to each other? Is there a formula for this?

You've got $n(n+1)(n+2)$ for $n=2000,\ldots,2009$. You want divisors that they all share in common.
Given three consecutive integers, exactly one of them is divisible by $3$; therefore their product is divisible by $3$. That is true of all $10$ of the products, so $3$ is a common divisor. Among the ten products, there are some, such as $2000\times2001\times2002$ in which $9$ is not a divisor. Therefore the highest power of $3$ that divides all ten products is $3$ itself.
If $n$ is even then so is $n+2$, and one of them is divisible by $4$ and the other is not, so the product is divisible by $8$. But if $n$ is odd, then so is $n+2$, and $n+1$ is even. Among the cases where $n+1$ is even, we have $n+1=2002$, which is not divisble by $4$, so $4$ is not a common divisor of all ten products. The highest power of $2$ that divides all ten products is therefore $2$ itself.
The next prime number is $5$. Some of these products are not divisible by $5$; for example $2001\times2002\times 2003$ is not. So $5$ is not a common divisor.
$7$ divides $2002$ and $2009$ and nothing between those, so $2003\times2004\times2005$ is not divisible by $7$.
The next prime is $11$. This one divides $2002$ but nothing else until you get up to $2013$, which is not included, so $11$ does not divide $2003\times2004\times 2005$. In a similar way you can rule out all larger primes as common divisors.
Therefore the g.c.d. is $2\times3=6$.
