What you have then is that

since all the cubed terms cancel from either side of your original equation.

Dividing by 3 will simplify this a little to

Now the trivial solution is obvious

The next step I would advise is setting one of the variables, let's say then we have

so we have a solution of the form

where is any non zero integer (otherwise it's a repeat of the trivial solution)

You can then use the same argument for but the solutions should be very similar (and hopefully obvious).

Finally assume all are non zero.

If all are positive then every term in the sum is positive and thus cannot be zero, similarly if all are negative. So we can assume that is negative and the other two are positive. If we let where is any positive integer we get

which simplifies to

which simplifies further to

Applying the quadratic formula to this will yield the solution

So we have another solution set and again the other two variables will yield similar solutions.

As far as I can tell this forms a complete set of solutions.

Hope this helps.