Suppose $3\
mid a$. Since $a^3+3ab=a^2(a+3b)$ and any prime $p$ dividing $a$ does not divide $a+3b$, we conclude that $a^2$ and hence also $a$ is a cube. Then all we need is that $a+3b$ is also a cube. So let $u$ be an arbitrary integer not divisible by $3$, let $k$ be an arbitry integer relatively prime to $u$. Then $a = u^3$, $b=\frac{(u+3k)^3-u^3}3=3ku^2+9k^2u+9k^3$, $c= u^2(u+3k)$ is a solution. (And all solutions with $3\
mid a$ are of this kind - why?)
Suppose $3\mid a$, say $a=3a_1$. Then $3\mid c$, say $c=3c_1$, and we are led to $a_1^2(a_1+b)=c_1^3$. Since $\gcd(a_1,a_1+b)=1$, we conclude again that $a_1^2$ (and hence $a_1$) is a cube and then that $a_1+b$ is also a cube. So let $u$ be an arbitrary integer, $k$ an integer relatively prime to $k$. Then $a= 3u^3$, $b=(u+k)^3-u^3=3k+3k^3+k^3$, $c= 3u(u+k)$ is a solution. (And all solutions with $3\mid a$ are of this kind - why?)