we can solve by fining GCD of 2 and 3 but here I shall solve it by a different approach
We can rewrite this as
$3y = 2x -8$
RHS is even so LHS should be even or y should be even
So let $y =2k$
We get $6k = 2x -8$
Or x = $3k + 4$
So $x = 3k + 4$ and $y = 2k$ are parametric form of solution
by choosing any integer for k we can get particular solution
k = 1 gives (7,2)
k =2 gives (10,4)
We can solve the 2 equations and get $a=5$ and $b=2$ and hence $a^3-b^3=5^3-2^3=117$
but the above approach is longer as compared to the approach below
We have
$(a-b)^3 = a^3-3a^2b + 3ab^3 - b^3 = a^3-b^3-3ab(a-b)$
or $a^3-b^3= (a-b)^3 + 3ab(a-b)$
Putting the values we get $a^3-b^3=3^3 + 3 * 10 * 3 = 117$
In this process we have avoided solving of equations
$16p+1$ is perfect cube so let is be $x^3$
So $x^3-1 = 16p$
Or $(x-1)(x^2+x+1) = 16p$
Now $x^2+x+1$ is is odd so $16| x-1$
So $(x-1) = 16m$ and $x^2+x+1 = n$ where m and n are odd
So p = mn
As p is odd $m = 1, n = p$ or $m = p, n= 1$
n= 1 gives $x^2+x+1 = 1 $ or x = 0 which is not possible
so m= 1 and x= 17 giving n = 307 wihich is a prime
so $p = 307$
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
