Le $P(a,b,c) = abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$
Let us first find out $a^3 \mod 7$
We have $a^3 \mod 7 \in \{1,-1\}$ when $a \mod 7 \ne 0$
if $a \mod 7 = 0$ or $b \mod 7 = 0$ or $c \mod 7 \ne 0$ we have $7 | P(a,b,c)$
$a^3 \mod 7 \in \{1,-1\}$
$b^3 \mod 7 \in \{1,-1\}$
$c^3 \mod 7 \in \{1,-1\}$
each can take one of two values and there are 3 differences so one of them has has to be zero
hence $7 | P(a,b,c)$
Proved
