$$a^3+b^3+c^3-3abc=6d^3$$ Since $(a,b,c)$ is symmetrical, let's assume $a<b<c$ and $d>0$.
Is there only one set of (nontrivial) nonzero integer solutions to this Diophantine equation? $$\left(a,b,c,d\right)=\left(-1,2,5,3\right)$$
Note that these solutions $\left(a,b,c,d\right)=\left(-8,-5,-5,-3\right),\left(4,7,7,3\right),\left(5,5,8,3\right)$ etc are not considered.
Above equation shown below:
x^3+y^3+z^3-3xyz=6d^3 ----(1)
In Individ solution take (a,b,k,t)=(98,21,3,5) we get
(x,y,z,d)=(184,190,214,42)
So the answer to "OP" question is negative. There
are multiple solutions to equation (1).