Nontrivial integer solutions of $\sum_{i=1}^3 a_i ^3=\sum_{i=1}^3 b_i ^3$ and $\sum_{i=1}^3 a_i =\sum_{i=1}^3 b_i$ 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.
 A: Some  solutions:
$$\matrix{a_1 &= 1,\; a_2 &= 5,\; a_3 &= 5,\;  b_1 &= 2,\; b_2 &=  3,\; b_3 &= 6\cr
a_1 &= 1,\; a_2 &= 9,\; a_3 &= 9,\;  b_1 &= 4,\; b_2 &=  4,\; b_3 &= 11\cr
a_1 &= 1,\; a_2 &= 10,\; a_3 &= 14,\;  b_1 &= 3,\; b_2 &=  7,\; b_3 &= 15\cr
a_1 &= 1,\; a_2 &= 10,\; a_3 &= 15,\;  b_1 &= 4,\; b_2 &=  6,\; b_3 &= 16\cr
a_1 &= 2,\; a_2 &= 8,\; a_3 &= 10,\;  b_1 &= 4,\; b_2 &=  5,\; b_3 &= 11\cr
a_1 &= 2,\; a_2 &= 10,\; a_3 &= 12,\;  b_1 &= 3,\; b_2 &=  8,\; b_3 &= 13\cr
a_1 &= 2,\; a_2 &= 10,\; a_3 &= 10,\;  b_1 &= 4,\; b_2 &=  6,\; b_3 &= 12\cr
a_1 &= 5,\; a_2 &= 10,\; a_3 &= 11,\;  b_1 &= 6,\; b_2 &=  8,\; b_3 &= 12\cr
}$$
EDIT: Here's a nontrivial infinite family of solutions (of course you can also multiply any solution by a constant factor).
$$a_1 = t, \; a_2 = 5 t, \; a_3 = 7 t - 2, \; b_1 = t + 1, \; b_2 = 5 t - 2,\; b_3 = 7 t - 1 $$
A: $$\left\{\begin{aligned}&X_1+X_2+X_3=Y_1+Y_2+Y_3\\&X_1^3+X_2^3+X_3^3=Y_1^3+Y_2^3+Y_3^3\end{aligned}\right.$$
$$X_1=(b+c-p)(a^2+a(b+c-p)+b(c-p))-cp(c-p)$$
$$X_2=(b+c-p)(a^2+a(b+c-3p)+b(c-p))+p(2p-2b-c)(c-p)$$
$$X_3=(b+c-p)(a^2-a(b+c-p)-b(c-p))+(2ab-2ap+cp)(c-p)$$
$$Y_1=(b+c-p)(a^2+a(c-b-p)+b(c-p))+(2bp-2b^2-cp)(c-p)$$
$$Y_2=(b+c-p)(a^2+a(b-c-p)+b(c-p))+c(p-2b)(c-p)$$
$$Y_3=(b+c-p)(a^2+a(b+c-p)+(b-2p)(c-p))+(2ab+cp-2ap)(c-p)$$
A: if you put
$\alpha_1=a_2+a_3$
$\alpha_2=a_1+a_3$
$\alpha_3=a_1+a_2$
$\beta_1=b_2+b_3$
$\beta_2=b_1+b_3$
$\beta_3=b_1+b_2$
Then your (3) becomes $\alpha_1\alpha_2\alpha_3=\beta_1\beta_2\beta_3$ and (2) becomes $\alpha_1+\alpha_2+\alpha_3=\beta_1+\beta_2+\beta_3$
Of course, some of the solutions of the system above give fractional $a_i,b_i$, since the determinant of the transformation is $2$. But if we get a fractional solution, we can multiply it by 2 to get an integer solution.
The link Two sets of 3 positive integers with equal sum and product will give you some examples of non-trivial $\alpha_i,\beta_i$
For example, the pairs $(8,12,15)$ and $(9,10,16)$ will generate the nontrivial solution of $(2.5,5.5,9.5)$ and $(1.5,7.5,8.5)$, which by doubling gives: $(5,11,19)$ and $(3,15,17)$. It would be interesting though if somebody would come with a closed-form formula to generate them.
A: Here is one non-trivial solution:
$$ 1 + 5 + 5 = 2 + 3 + 6 $$
$$ 1^3 + 5^3 + 5^3 = 2^3 + 3^3 + 6^3 $$
The first line sums are $11$ and the second line sums are $251$.
