Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is? 
Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is?

The correct answer is $\dfrac13$ while mine is $\dfrac{445}{2003}$
My attempt:
For $x^3+y^3$ to be divisible by $3$, EITHER both $x$ and $y$ should be a multiple of $3$ OR one of them should leave remainder $1$ when divided by $3$ and the other should leave remainder $2$.
Therefore, $$\text{no. of ways} = \frac{668 \times 667 + 668\times 668}{2004\times 2003} = \frac{445}{2003}$$ 

It is immediate  (by using arithmetic progression) to get that there are equally $668$ numbers of residues $0,1,2$ modulo $3$. We have
$$(3n)^3+(3n)^3\equiv 0\pmod3\\(3n)^3+(3n+1)^3\equiv 1\pmod3\\(3n)^3+(3n+2)^3\equiv 2\pmod3\\(3n+1)^3+(3n+1)^3\equiv 2\pmod3\\(3n+1)^3+(3n+2)^3\equiv 0\pmod3$$
It follows$$\binom{668}{2}+668^2=669002\space\space\text{favorable cases.}\\\binom{2004}{2}=2007006\space\space\text{possibilities}$$
Thus the probability is $$\frac{669002}{2007006}=\frac13$$
