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Multiplying both sides by $17(x^3+y^3)$ , we get

$x^3+y^3=17(289-3xy)$

$x^3+3(17)(xy)+y^3 = 17^3$

Now, notice that if $x+y = 17$ , then $(x+y)^3 = x^3+3(x+y)(xy)+y^3 = x^3+3(17)(xy)+y^3 = 17^3$ , so we have our case $x+y = 17$ , giving us $18$ solutions.

Also, notice that since this is a symmetric equation in $x$ and $y$ , we will have solutions for $x = y$ . Plugging this in the original equation gives us $2x^3+3(17)(x^2) = 17^3$ , which yields $x=y=-17$ , for another solution.

So, we have a total of $18+1 = \boxed{19}$ solutions.

Multiplying both sides by $17(x^3+y^3)$ , we get

$x^3+y^3=17(289-3xy)$

$x^3+y^3-17^3-(3)(xy)(-17)=0$

Note that $a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$

Setting $a=x$ , $b=y$ , and $c=-17$ , we have

$x^3+y^3-17^3-(3)(xy)(-17)=\frac{1}{2}(x+y-17)[(x-y)^2+(y+17)^2+(-17-x)^2]$

$0=\frac{1}{2}(x+y-17)[(x-y)^2+(y+17)^2+(-17-x)^2]$

We have either $x+y=17$ or $(x-y)^2+(y+17)^2+(-17-x)^2=0$ , which, by following the restrictions of the problem, give $(0,17),(1,16),(2,15)...(17,0)$ and $(-17,-17)$ , a total of 19 solutions.

This problem was adapted from the 1999 American High School Mathematics Examination (AHSME) .