Is $A$ a perfect square? Consider
$$A=a^2+2ab^2+b^4-4bc-4b^3,$$
where $a,b,c\in\mathbb{Z}$ and $b\neq0$ such that $b|a$ and $b|c$, so $b|A$.
Now I want to know that Is $A$ a perfect square?

No, $A$ is not a perfect square for such $a,b,c$ values. As a counterexample, take $a=b=c=1$, for which $A=-4$.

EDIT: The answer below corresponds to the original question "Can $A$ be a perfect square?".
Yes, it can be 0. Let $a=bn$ and $c=bm$. Then:
$$\begin{align}
A&=a^2+2ab^2+b^4-4bc-4b^3\\
&=b^2n^2+2b^3n+b^4-4b^2m-4b^3\\
&=b^2 \big((n+b)^2-4(m+b)\big)
\end{align}$$
So any values that make $(n+b)^2-4(m+b)$ a square are valid ones. So, for instance, any $n=-b$ and $m=-b$ are enough, but you can find many others.
