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

Since $57$ is odd, we need $x+y+z$ odd.
Since $3\mid 57$ and $n\equiv n^3\bmod 3$, we need $3\mid x+y+z$.
Given that $-5\le x,y,z\le5$, it's relatively quick to eliminate $x+y+z=\pm 9$ and identify that we need $x+y+z=\pm 3$.
Then taking $x\ge y\ge z$ we must have $x\ge 3$ initially and then after considering a couple of cases we get $x\ge 4$, and we can also quickly eliminate $x=5$ leaving only $x=4$ to explore, which quickly leads to $y=1, z=-2$.
If all permutations are required then there are $3!=6$ arrangements of these values, of course.
