The equation $a^3+b^3+c^3=kabc$ I am interested in the equation $a^3+b^3+c^3=kabc$ for $a,b,c \in \mathbb{N}$. We have by AM-GM:
$$a^3+b^3+c^3 \geqslant 3abc \implies k \geqslant 3$$
Since $k=3$ is the equality case, the solutions for $a^3+b^3+c^3=3abc$ is $(a,b,c)=(x,x,x)$ for some $x \in \mathbb{N}$. However, it is not clear whether it is possible to solve any case $k>3$, atleast in an elementary fashion. 
For which values of $k$ has this equation been solved? Are there any results for any $k>3$ where $k \in \mathbb{Q}$ (or specifically in $k \in \mathbb{N}$)? 
EDIT : I must specify that as the equation is homogeneous, it is obvious that you can generate a family of solutions from a primitive solution by scaling. Thus, I consider only the cases where $\gcd(a,b,c)=1$. It can easily be seen that this also means they are pairwise relatively prime.
I am aware that there are 'some' solutions for specific $k$. This doesn't answer my question. I am looking for characterization of all primitive solutions, generating infinitely many primitive solutions, proving infinitude of primitive solutions, non-existence of solutions etc. for specific $k$.
 A: We can get solutions one by one through seeking rational roots of a cubic equation.
Assume wlog $a\le b\le c$.  Pick values of $a$ and $b$ that meet the above ordering requirement.  Then render a cubic equation for $c$:
$c^3-(kab)+(a^3+b^3)=0$
And solve the original equation for $k$:
$k=(a^3+b^3+c^3)/(abc)$
We then have a solution if $c$ divides $a^3+b^3$ and $abc$ divides $a^3+b^3+c^3$.
Suppose, for instance, $a=b=1$.  Then $c|2$ by the first criterion.  We find that the second criterion also holds for each candidate $c=1$ and $c=2$ giving two solutions $(a,b,c,k)=(1,1,1,3)$ and $(a,b,c,k)=(1,1,2,5)$.
Now try $a=1, b=2$. Here $c\in\{1,3,9\}$ by the first criterion, but we cover $c=1<b$ with a smaller ordered pair for $(a,b)$.  For $c=3$ we infer $(a,b,c,k)=(1,2,3,6)$ and for $c=9$ we succeed with $(a,b,c,k)=(1,2,9,41)$.
A: Equation, $a^3+b^3+c^3=k(abc)$
Has parametric solution:
$k=(m^2+8)$
$a=m(m^2+3)$
$b=(m^2+4m+7)$
$c=(4m-m^2-7)$
For, $m=2$,we get,$(a,b,c,k)=[(14),(19),(-3),(12)]$
For, $m=3$,we get,$(a,b,c,k)=[(9),(7)(-1),(17)]$
