The reason is (partially) given in the part of the sentence which was not quoted:
since every cube is congruent to $0$ or $\pm1 \pmod 9$
More specifically, if $x\equiv a\pmod 3$, where $a$ is $-1,0$ or $1$, then $x^3\equiv a\pmod 9$. This is a very useful restriction, especially if $k\equiv\pm3\pmod 9$.
In particular, if $k = x^3 + y^3 + z^3$ and $k\equiv 3\varepsilon\pmod 9$ with $\varepsilon=\pm1$, the only way to solve the original equation is for all three of $x^3,y^3,z^3$ to be congruent to $\varepsilon\pmod 9$, which means $x,y,z$ are each congruent to $\varepsilon \pmod 3$. Thus we can have $3 \equiv 1 + 1 + 1 \pmod 9$, but it cannot be produced by any other combination of three $-1$s, $0$s and $1$s; similarly for $-3$, which is the relevant value for $k = 33$.