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Algebra Difficulty 4.9 AIME Find the answer

Compute: ⌊200532003⋅2004−200332004⋅2005⌋\left\lfloor\frac{2005^{3}}{2003 \cdot 2004}-\frac{2003^{3}}{2004 \cdot 2005}\right\rfloor

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let x=2004x=2004. Then the expression inside the floor brackets is (x+1)3(x−1)x−(x−1)3x(x+1)=(x+1)4−(x−1)4(x−1)x(x+1)=8x3+8xx3−x=8+16xx3−x\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} Since xx is certainly large enough that 0<16x/(x3−x)<10<16 x /(x^{3}-x)<1, the answer is 8.

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