<s>[INST] Solve $x^3 +y^3 + z^3 =57$ How can we solve $x^3 + y^3 + z^3 =57$ efficiently in a shorter way. $x$ $y$ and $z$ are integers. Given that modulus of $x$ $y$ and $z$ is less than or equal to five. We can of course do by hit and trial but what is the method of solving such questions. I actually stumbled upon this equation while solving a determinant. How to proceed. Pls help
 [/INST] With some forethought, you can bring it down to one case to check directly, which turns out to be a solution. A previous version of this answer examined the equation mod $7$, which led to $16$ cases to directly examine. This answer examines mod $9$, which works out even better. (The reason $7$ and $9$ are good moduli to consider is because there are relatively few cubes mod these numbers.) 
Mod $9$, the only cubes are $0$, $1$, and $8$. For solutions to $X+Y+Z\equiv57\equiv3$, the only solution is $1+1+1\equiv3$. This means $x^3\equiv y^3\equiv z^3\equiv1$ mod $9$, which means $x\equiv y\equiv z\equiv1$ mod $3$. So all of $x$, $y$, and $z$ are among $\{-5,-2, 1, 4\}$.
As in @RossMillikan's answer, we can assume $x, y\leq z$ and $z$ must equal $4$. So now we must solve $x^3+y^3=-7$, with $x\leq y$ among $\{-5,-2, 1, 4\}$. 
Consider this equation mod $7$, where the cubes of $\{-5,-2, 1, 4\}$ are equivalent to $\{1,6, 1, 1\}$ respectively. The only way to sum two of these and get $-7\equiv0$ is using $6$ and $1$. Therefore one of $x,y$ is $-2$. Wlog, assume it's $x$, and examining $(-2)^3+y^3=-7$, we have that $y=1$ makes a complete solution $(x,y,z)=(-2,1,4)$.
</s>