3. Compute:
$$
\left\lfloor\frac{2005^{3}}{2003 \cdot 2004}-\frac{2003^{3}}{2004 \cdot 2005}\right\rfloor .
$$
Solution:
8
Let $x=2004$. Then the expression inside the floor brackets is
$$
\frac{(x+1)^{3}}{(x-1) x}-\frac{(x-1)^{3}}{x(x+1)}=\frac{(x+1)^{4}-(x-1)^{4}}{(x-1) x(x+1)}=\frac{8 x^{3}+8 x}{x^{3}-x}=8+\frac{16 x}{x^{3}-x} \text {. }
$$

Since $x$ is certainly large enough that $0<16 x /\left(x^{3}-x\right)<1$, the answer is 8 .