I'm not going to do all the calculations, just get things started.
From $a^3+40033 = d = b^3+39312$, one gets $b^3-a^3=721$, or $(b-a)(b^2+ab+a^2)=7\cdot103$, so $b-a\in 1,7,103,721$. You can do these cases one at a time. For example, if $b-a=1$ then you have $(a+1)^2+a(a+1)+a^2=721$, or $a^2+a-240=0$, which has no solutions. If one of the cases does produce a solution, then you can compute the corresponding $d$ and check it against the equation $d=c^3+4104$.
There may well be a slicker approach, but this should work.
Added 5/22/13: I just discovered a mildly embarrassing error in my answer. The equation $a^2+a-240=0$ does have solutions: The quadratic factors as $(a-15)(a+16)$. (I had mentally multiplied $4\times240=920$ and knew that a discriminant of $921$ was too close to $900$ to be a perfect square.) I finally caught my error when I decided to try to make my approach a little slicker:
If you write $b=a+k$, then $b^3-a^3=721$ becomes $k(k^2+3ka+3a^2)=7\cdot103$, so again $k \in 1,7,103,721$. But now you can immediately rule out $k=103$ and $k=721$, because the quadratic factor $k^2+3ka+a^2$ would obviously produce a number way too large (since $a$ is also required to be positive). So that leaves $k=1$ and $k=7$, both of which do lead to solutions. It so happens, though, that they lead to $(a,b,c)=(15,16,34)$ and $(2,9,33)$, respectively, which the OP explicitly disallowed.