How to Find the all integer solutions for:
We have $x^3+y^3=3-z^3$ and $x+y=3-z$. Since $x+y$ divides $x^3+y^3$, we conclude that $z-3$ divides $z^3-3$, and therefore $z-3$ divides $24$. Similar considerations apply to $x$ and $y$.
So we are down to a finite and indeed fairly short list of candidates. We can use further little tricks to winnow the list.
Remark: Let's throw in some number theory. It is a sometimes useful fact that $a^3$ is always congruent to $0$, $1$, or $-1$ modulo $9$. Thus if $x^3+y^3+z^3=3$, we must have $x^3$, $y^3$, and $z^3$ all congruent to $1$ modulo $9$. It follows that all of $x$, $y$, and $z$ are congruent to $1$ modulo $3$, and hence so are $x-3$, $y-3$, and $z-3$. The only divisors of $24$ that satisfy this condition are $1$, $4$, $-2$, and $-8$. So our only candidates for $x$, $y$, and $z$ are $4$, $7$, $1$, and $-5$.
I won't give you the solution, but a path. First find all solutions where all numbers are >= 0, that's very easy. In other cases, you have both positive and negative numbers. Show why you can't have x = -y, for example. And then you show that the largest number cannot be very large.