Solve $a^3 + b^3 + c^3 = 6abc$ Find solutions for  $a^3 + b^3 + c^3 = 6abc$ in  $\mathbb{N}$, such that $gcd(a,b,c) = 1$, except for $(1,2,3)$ and its permutations.

Using trial and error I found out that if $a,b,c$ are solution of the equation, then they are in arithmetic progression. I've managed to prove that conjecture, assuming that $c>b>a$ and let $k$ be their common difference in the arithmetic progression. Then WLOG we have:
$$b = c-k \quad \quad a = c-2k$$
Now the equation looks like:
$$(c-2k)^3 + (c-k)^3 + c^3 = 6(c-2)(c-1)c$$
After expanding we have:
$$c^3 - 6kc^2 + 12ck^2 - 8k^3 + c^3 - 3kc^2 + 3ck^2 -k^3 + c^3 = 6c^3 - 18kc^2 + 12ck^2$$
$$3c^3 - 9kc^2 + 15ck^2 - 9k^3 = 6c^3 - 18kc^2 + 12ck^2$$
$$c^3 - 3kc^2 + 5ck^2 - 3k^3 = 2c^3 - 6kc^2 + 4ck^2$$
$$-c^3 + 3kc^2 + ck^2 - 3k^3 = 0$$
Now it's easy to see that if $k=c$, then the LHS will be zero, so one of the zeroes of the polynomial is $c_1 = k$, now factorizing we have:
$$(c-k)(3a^2 + 2ax - x^2) = 0$$
$$(c-k)(c+k)(c-3k) = 0$$
Now we have three distinct cases:
Case 1: $c = k$
This implies that $b = 0$ and $a = -k$. But because $k \in \mathbb{N}$, both $a,b \not\in \mathbb{N}$, violating the initial conditions.
Case 2: $c = -k$
Obviously the initial condition is already violated, becasue $k \in \mathbb{N}$, so from the relation $c \not\in \mathbb{N}$
Case 2: $c = 3k$
This implies that $b = 2k$ and $a = k$. Now we have one 3-tuple $(3k,2k,k)$ and it's permutation as solution, where $k \in \mathbb{N}$. But it's easy to note that $k$ is a common factor for $a,b,c$ so we have:
$$gcd(a,b,c) = k$$
But because we want $gcd(a,b,c) = 1$, this implies that $k=1$, which means we have only one solution for  $a^3 + b^3 + c^3 = 6abc$ in  $\mathbb{N}$, such that $gcd(a,b,c) = 1$ and it $(1,2,3)$, solution that is already given.
Now my question is what I'm missing. Is there really no other solutions such that $gcd(a,b,c) = 1$? Or maybe there is a different way to obtain solution except for my method using arithmetic progression?
 A: It's easy to prove that at least one of the variables needs to be an even number. We know that:
$$6|(n-1)n(n+1)$$
Because in three consecutive numbers, one is divisible with three and at least one is divisible with 2. So we have:
$$6|n^3 - n$$
$$n^3 = n \pmod 6$$
Now we have:
$$a^3 + b^3 + c^3 \equiv a + b + c \equiv 0 \pmod 6$$
Beacuse the modulo is an even number that means that the sum $a+b+c$ is an even number also. We know that the sum of 3 odd numbers will be odd number, so it's impossible $a,b,c$ to be odd number, because there won't be solution. So it means that at least one of the variables is an even number.
WLOG we can set $b=2k$. Now we can continue:
$$b-2k = 0$$
Now we can multiply both sides with $b(b+2k)$. Note that won't give another solution, because it'll imply that b is $0$ or a negative number, which violate the condition. Now we have:
$$b(b+2k)(b-2k) = 0$$
$$b(b^2 - 4k^2) = 0$$
$$b^3 - 4bk^2 = 0$$
$$3b^3 - 12bk^2 = 0$$
$$6b^3 - 6bk^2 = 3b^3 + 6bk^2 = 0$$
$$6b(b^2 - k^2) = (b^2 - 3kb^2 + 3bk^2 - k^3) + b^3 + (b^3 + 3kb^2 + 3bk^2 + k^3) = 0$$
$$6b(b-k)(b+k) = (b-k)^3 + b^3 + (b+k)^3$$
Now if we substitute WLOG:
$$b+k=c \quad \quad b-k=a$$
$$6abc = a^3 + b^3 + c^3$$
Because $b,k \in \mathbb{N}$ it means that also $a,c \in \mathbb{N}$. So this proves that for any $b=2k$, there are integer solutions, such $a=k$ and $c=3k$.
But because $k$ is a factor of all of them it's easy to see that:
$$gcd(a,b,c) = gcd(k,2k,3k) = k$$
Because we want $gcd(a,b,c) = 1$, that implies that $k=1$ and that the only primitive solution of this equation is $(1,2,3)$ and its permutation. 
