The goal is to understand a set of nontrivial solutions of cubic polynomials
$$ \sum_{i=1}^3 a_i ^3=\sum_{i=1}^3 b_i ^3 \Rightarrow a_1^3+a_2^3+a_3^3=b_1^3+b_2^3+b_3^3, \tag{1} $$ $$ \sum_{i=1}^3 a_i =\sum_{i=1}^3 b_i \Rightarrow a_1+a_2+a_3=b_1+b_2+b_3 , \tag{2} $$ for $a_i,b_i\in \mathbb{Z}$ where we demand $(a_1,a_2,a_3)\neq (b_1,b_2,b_3)$ or any of its permutation (total $3!=6$ cases).
Question: Are there any nontrivial solutions? (e.g. $a_i,b_i\in \mathbb{Z}$ where we demand $(a_1,a_2,a_3)\neq (b_1,b_2,b_3)$ and other $3!$ permutations.) How many simple but nontrivial solutions are there in the smaller values of $a,b,c,d$? (The simple form the solutions are the better.) Or can one prove or disprove that there are nontrivial solutions? (Even better, if we can get a quick algorithm to generate the solutions.)
A non-trivial example or a proof of non-existence is required to be accepted as a final answer.
Note add:
My trials/attempts: Since it is encouraged to show one's own attempt.
First, it looks like if we have only $i=1,2$ instead of summing over $i=1,2,3$, we have $\sum_{i=1}^2 a_i ^3=\sum_{i=1}^2 b_i ^3 $ and $\sum_{i=1}^3 a_i =\sum_{i=1}^3 b_i$, the nontrivial solutions seem to be impossible (?), but no rigorous proof has be shown yet.
Let me share a few comments on what these two equations boil down to:
If we take (2)$^3$-(1), we get $$ (a_1+a_2)(a_1+a_3)(a_2+a_3)=(b_1+b_2)(b_1+b_3)(b_2+b_3), \tag{3} $$ we can further use (2) and (3) to get a lot more interesting constraints. However, I haven't got any other linear relation constraint as useful as (2) yet.