Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$. 
Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$.

What I Tried:- Given this expression I had no idea to claim something out of it, because it looks like we need to $437$ in terms of sums of $2,3$ (the $6$ does not matter much here). I could break $(8n^3 + m^3)$, but it didn't help much.
Instead I simplified the whole expression and got :-
$$2m^3 + 6m^2 + 6m + 16n^3 - 36n^2 + 27n = 437.$$
$$\rightarrow 2m^3 + 6m^2 + 6m + 16n^3 - 36n^2 = 437 - 27n.$$
From here, $2 \vert (437 - 27n)$, so $n$ is odd. Let $n = (2k + 1)$. Substitute this back to the equation to get :-
$$64k^3 + 24k^2 + 3k + m^3 + 3m^2 + 3m = 215.$$
Now I am stuck here, it is given $(k,m)$ are integers, so they can range to negative values also. So I was not able to make an upper bound on $k$, and I have no other idea what I can claim from here.
Can anyone help me? Thank You.
 A: First, substitute $x=m+1$ and move the constant term to the left side:
$$2x^3+16n^3-36n^2+27n=439$$
Then, note that $n$ must be odd.  We can replace $n=2y+1$ and divide out by $2$ and get:
$$x^3+64y^3+24y^2+3y=216$$
Some algebraic manipulation (multiplying both sides by $8$, adding $1$ to both sides, factoring, and then replacing $z=8y+1$):
$$8x^3+512y^3+192y^2+24y=1728$$
$$8x^3+512y^3+192y^2+24y+1=1729$$
$$(2x)^3+(8y+1)^3=1729$$
$$(2x)^3+z^3=1729$$
Expressing $m$ and $n$ in terms of $x$ and $z$ gives $m=x-1$, $n=\frac{z-1}{4}+1$.
$1729$ is famously the Ramanujan-Hardy taxicab number, the smallest natural number that can be written as the sum of two cubes in two different ways.  $1729=1^3+12^3=10^3+9^3$.  This results in possible $(x,z)$ of $(6,1)$ or $(5,9)$.  In terms of $(m,n)$, we have $(4,3)$ and $(5,1)$.
Therefore, our possible values for $mn$ are $5$ and $12$.
A: Edit: This is similar Moko19's answer (written in paralel but posted too late...), but I keep it here as it describes process how to get the solutions for the $a^3+b^3=1729$.
Notice that $m^3+3m^2+3m=(m+1)^3-1$, and similarly $$64k^3 + 24k^2 + 3k=\frac{1}{8}((8k)^3+3(8k)^2+3(8k))=\frac{1}{8}((8k+1)^3-1).$$
And so
$$
\frac{1}{8}((8k+1)^3-1)+(m+1)^3-1=215
$$
which in turn after simplification gives
$$
(8k+1)^3+(2m+2)^3=1729.
$$
Let $a=8k+1=4n-3$ and $b=2m+2$. Now notice that $a^3+b^3=(a+b)(a^2-ab+b^2)=1729$ and also $a^2-ab+b^2>0$, which implies $a+b>0$ (regardless of whether any of $a,b$ is negative). Finally, we need to only check finitely many cases, namely for each positive divisor $d \mid 1729 = 7\cdot 13\cdot 19$ (so $d\in \{1, 7, 13, 19, 91, 133, 247, 1729\}$), let $a+b=d$ and $a^2-ab+b^2=\frac{1729}{d}$ , which simplifies to quadratic (diophantine) equation in one variable.
For example, for $d=13$ we have $a+b=13$ and $a^2-ab+b^2=133$, and hence substituting first into the second and cancel $3$, we have $a^2-13a+12=0$. This solves to $a=12$ or $a=1$, applying the original substitution backwards we get $n=1,m=5$. Same way, we find that only other value that works is $d=19$, leading to $n=3,m=4$.
Bonus: You might recognize a famous Ramanujan's taxicab number 1729, which is the smallest natural number expressible as a sum of positive(!) two cubes in exactly two different ways: $$1729=1^3+12^3=9^3+10^3.$$
Which also gives us a solution (although we still need to be careful about the negative cubes).
A: This is just to fill in the unfinished part of Moko19's answer, showing that there are no solutions with mixed signs. That is, we'll show here that $1729$ can only be written as the sum of two positive cubes.
If $a^3-b^3=1729$ with $a\gt b\gt0$, then we have
$$1729=(a-b)(a^2+ab+b^2)=(a-b)((a-b)^2+3ab)\gt(a-b)^3$$
Consequently, since $(a-b)\mid1729=7\cdot13\cdot19$, we can only have $a-b=1$ or $a-b=7$. If $a-b=1$, then we have $ab=(1728-1)/3=576=24^2$, while if $a-b=7$, then we have $ab=(247-49)/3=66=11\cdot6$. It's easy to see that neither of these is possible.
