$6(m + 1)(n − 1), 6 + (m − 1)(n + 1)$ 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.
Hint: If I am not mistaken, we have$$m=\frac{24-a^3+6b^3}{12} $$
$$n=\frac{12-a^3+18b^3-12c^3}{12} $$
Hint: If I am not mistaken, we have$$m=\frac{24-a^3+6b^3}{12} $$
$$n=\frac{12-a^3+18b^3-12c^3}{12} $$