Four integers that satisfy $a+b+c+d\; =\; -3$ and $a^{3}+b^{3}+c^{3}+d^{3}\; =\; 3$ 
Find a set of 4 integers that satisfy $$a+b+c+d\; =\; -3$$ and $$a^{3}+b^{3}+c^{3}+d^{3}\; =\; 3$$

I am really not sure how to proceed. I tried letting $d = -c$ to see if that would yield a possible solution but it did not work. Further, trying to cube the first equation just yields a big mess. I think there might be a way to factor using sum of cubes and make substitutions that could simplify it but I don't have any insight as to how to approach it...
 A: Well, there is a solution with $a=b=c$.
EDIT:
With $c = -a + k$ and $d = -b-k-3$, the system becomes 
$$ k a^2  - k^2 a  + 3 k^2 + 9 k + 10 = (k+3) b^2 + (k+3)^2 b $$
which has infinitely many integer solutions for some integer values of $k$.
For example, with $k = 2$, writing $a = A/4 + 1$ and $b = (B-5)/2$ yields
$$   A^2 = 10 B^2 + 86 $$
(where we want solutions where $A$ is divisible by $4$).  One solution is $A = 24, B = 7$ (corresponding to $a = 7, b = 1$).  From the theory of
Pell equations, given one solution $(A,B)$ of $A^2 = 10 B^2 + 86$, another is
$(A' = 19 A + 60 B, B' = 6 A + 19 B)$.  In terms of $a$ and $b$, the mapping
is $$(a,b) \to (a',b') = (19 a + 30 b + 57, 12 a + 19 b + 33)$$
Repeating the mapping gives us an infinite family of solutions
$$ \eqalign{ [7, 1,& -5, -6]\cr
[220, 136,& -218, -141]\cr
[8317, 5257,& -8315, -5262]\cr
[315790, 199720,& -315788, -199725]\cr
[11991667, 7584193,& -11991665, -7584198]}$$
etc.
A: Those work, as do
-8,-8,3,10 and -6,-5,1,7
A: $(3, -2, -2, -2 ) $ works. Trial and error.
