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?

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.
