If $a$, $b$, $c$ are three positive integers such that $a^3+b^3=c^3$ then one of the integer is divisible by $7$ Let on contrary that none of the $a$, $b$, $c$ is divisible by $7$. Then either $a^3\equiv  b^3\pmod{7}$ or $b^3\equiv  c^3\pmod{7}$ or $c^3\equiv  a^3\pmod{7}$.
Now how to go further?
 A: If none of a,b,c are divisible by 7,
then each of a$^3$, b$^3$ and c$^3$ = either 1 or -1 (mod 7).
A contradiction ensues.
A: For any integer $x$ we have $x^3 \equiv -1 \text{ or } 0 \text{ or }1\pmod{7}$. 
$a^3+b^3=c^3$
$a^3+b^3+(-c)^3=0 \equiv 0\pmod{7}$ [Take modulo 7 on the whole equaton]
This sum can be zero iff one of them is $-1 \bmod7$, one is $1 \bmod 7$ and one is $0 \bmod 7$. Or if each one of them is $0 \bmod 7$.
The result follows.
A: There's 36 combinations of $a$ and $b$ in the integers mod 7, where they're not already $0$. Just try checking them.
A: $7$ is prime. This implies that the multiplicative group $\mathbb{Z}_7^*$ is cyclic and of order $6$. It follows that the image of the action $x\mapsto x^3$ is a subgroup of order $6/3=2$. So there are only two values for $x^3$ mod $7$.
(Plus a third value, $0$, since $0$ was excluded from $\mathbb{Z}_7^*$. But we are assuming none of the three variables are $0$ mod $7$.)
It's easy to see those values are $1$ and $-1\equiv6$. Now, can you solve this equation mod $7$: $$A+B\equiv C$$ using only the values $1$ and $-1$?
In general, this fact about how powers of integers mod $p$ can sometimes only take certain values can be useful. For example, consider $a^5+b^5=c^5$ and look at $11$ in place of $7$. It's the same situation, since $5$ divides $11-1$.
Or consider $a^4+b^4=c^{12}+7$ If you look at that one mod $13$, the $12$th power can only be $0$ or $1$. But the $4$th powers are so restricted that they cannot sum to $7$ or $8$.
