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?

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. 
