Given $x,y,z\in \mathbb{Z}$ such that $x^3+y^3-z^3$ is multiple of $7$ prove that at least one is divisible by $7$ 
Given $x,y,z\in \mathbb{Z}$ such that $x^3+y^3-z^3$ is multiple of $7$ prove that at least one is divisible by $7$

Given $n\in \mathbb{Z}$ then $n\in \lbrace [0],[1],[2],[3],[4],[5],[6]\rbrace$, of the same way
$n^3\in \lbrace[0],[1],[6] \rbrace$.
Since $7\mid x^3+y^3-z^3$ then $x^3+y^3-z^3\in [0]$ there be two possibilities
$x^3,y^3,z^3\in [0]$ and then $x,y,z\in[0]$ and the result is true.
$x^3\in [6],y^3\in [1],z^3\in[0]$ and then $z\in [0]$ therefore the results follows.
$x^3\in [0], y^3 \in [1], z^3 \in [1]$ and then $x\in [0]$and the afirmation is too true
The other possibilities only switch the order of the variables with the classes.
 A: If it's a more optimized proof you're looking for, these exist. Again, we notice that
$$a^3=a^{(7-1)/2}\equiv\left(\frac a7\right)\in\{-1,0,1\}\pmod7,$$
which we can also see by direct casework. We note that $a^3\equiv0$ if and only if $a\equiv0.$
The advantage of saying $\{-1,0,1\}$ is that we can say the sum $x^3+y^3-z^3$ lives in $[-3,3],$ and further, if all are nonzero, then this sum is $1+1+1\equiv1\pmod2$; i.e., the sum is odd. In particular, if all are nonzero, then $x^3+y^3-z^3$ is not divisible by $7,$ and we finish by contraposition.
A: By taste, I generally prefer solutions based on structural properties given implicitly in  the hypotheses. Here the natural structure is that of the field $\mathbf F_7=\mathbf Z/7\mathbf Z$, more precisely of its multiplicative group $\mathbf F_7^*$. From the properties of finite fields, we know that  $\mathbf F_7^*$ is cyclic of order 6 , hence $\cong  C_2 \times C_3$, where $C_2$ is generated by $-1$ and $C_3$ by $\zeta$, a 3-rd root of unity. In the given cubic equation, we may suppose that no variable is a multiple of 7, otherwise there is nothing to prove. Reduction modulo 7 then gives $a^3 + b^3=1$ in $\mathbf F_7^*$. The previous decomposition of $\mathbf F_7^*$ shows that the components of $a$ are $\pm 1$ and a power of $\zeta$, hence $a=\pm 1$, and the same for $b$. The  final equation mod 7 thus reads $(\pm 1) + (\pm 1) =1$, a contradiction.
This argument may seem circuitous, but I don't think so because: - first it avoids case by case verification, hence there is no risk of a possible omission  - second it appeals only to internal structural properties (e.g. no ordering relation was used), so it could be generalized to other analogous situations.
Addendum : To my surprise, I have just realized that the above argument proves (a weakened version of) Sophie Germain's result on the so called "first case" of Fermat's last theorem: if $p$ and $q= 2p+1$ are both prime (then $p$ is called a Sophie Germain prime), then for all integer solutions of $x^p + y^p = z^p$, $p$ must divide $xyz$ (see e.g. Wikipedia). This certainly cannot be checked case by case.
