<s>[INST] $3ab + a^3 - 2b^3 - 4a + 5b - 7 = 0$ I came across this problem:

Prove there arent't any $a$, $b$ integers that satisfy
  equation $3ab + a^3 - 2b^3 - 4a + 5b - 7 = 0$

Firstly, I've thought something like this:
$$(a^3 + b^3)-3b^3 + 3ab - 4a + 4b - 4 + b = 3$$
$$(a+b)(a^2-ab+b^2) - 3(b^3 - ab) -4(a-b+1)+b = 3$$
I think it's just a tricky exercise, but I can't find a way to solve it.
Some help would be apreciated.
Thanks!
 [/INST] $3ab + a^3 - 2b^3 - 4a + 5b - 7 = 0$
Consider the equation mod $3$.  Since $x^3\equiv x\pmod 3$ for all $x$, we have
$3ab+a-2b-4a+2b-7\equiv 0$ which is the same as $2\equiv 0\pmod 3$.
Contradiction.
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