$6(m + 1)(n − 1), 6 + (m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes. 
Find all pairs of integers (m, n) such that the integers $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.

I assumed $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ to be $a^3,b^3$ and $c^3$ respectively and deduced that $6(2mn+4)=a^3+6b^3,$ $12(n-m)=a^3-6b^3+36$ but
I can't solve it further. Please help.
 A: Suppose $m$ and $n$ such that the expressions are perfect cubes. Let $a$, $b$ and $c$ be integers such that
\begin{eqnarray*}
a^3&=&6(m+1)(n-1)&=&6mn-6m+6n-6,\\
b^3&=&6+(m-1)(n+1)&=&mn+m-n-6,\\
c^3&=&(m-2)(n+2)&=&mn+2m-2n-4.
\end{eqnarray*}
Then comparing coefficients shows that
$$a^3-18b^3+12c^3=90,$$
so in particular $a$ is divisible by $6$, and so plugging in $a=6A$ then yields
$$36A^3-3b^3+2c^3=15,$$
which shows that $c$ is divisible by $3$, and so plugging in $c=3C$ then yields
$$12A^3-b^3+18C^3=5.$$
Now reducing mod $9$ shows that
$$3A^3-b^3\equiv5\pmod{9},$$
and so $A^3\equiv-1\pmod{9}$ and $b^3\equiv1\pmod{9}$, so $A\equiv2\pmod{3}$ and $b\equiv1\pmod{3}$. Then plugging in $A=3\alpha+2$ and $b=3B+1$ yields
$$81\alpha^3+54\alpha^2+36\alpha-B^3-3B^2-3B+18C^3=-2.$$
Reducing mod $3$ shows that $B^3\equiv2\pmod{3}$ and so plugging in $B=3\beta+2$ shows that
$$27\alpha^3+18\alpha^2+12\alpha-9\beta^3-27\beta^2-27\beta+6C^3=8.$$
The left hand side is divisible by $3$ but the right hand side is not, a contradiction. Hence no such integers $m$ and $n$ exist.
A: Hint: If I am not mistaken, we have$$m=\frac{24-a^3+6b^3}{12} $$
$$n=\frac{12-a^3+18b^3-12c^3}{12} $$
