Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$. 
Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.

If $ab \mid (a+b)^²$, then $ab\mid a^2+2ab+b^2 \Longrightarrow ab\mid a^2, ab\mid 2ab$ and $ab\mid b^2$ right?
So since $(a-b)^2 = a^2-2ab+b^2$ from the assumption we have that $ab \mid a^2$ and $ab \mid b^2$. Now only remains to show that $ab \mid -2ab$ which is clearly true.
Is this valid? I'm not sure about the implication that $ab$ would divide all the terms in $a^2+2ab+b^2$.

We have $2 \mid 6=3+3$ but $2 \not \mid 3$, so the fact that if $x \mid y +z$ then $x \mid y$ and $x \mid z$ isn't valid (at least in general).
What you just have to use here is the fact that if
$$
x \mid y \text{ and } x \mid z
$$
therefore
$$
x \mid \lambda y + \mu z.
$$
Here, if $ab \mid (a+b)^2$ and since it's obvious that $ab \mid ab$, we have
$$
ab \mid (a+b)^2 -4ab=(a-b)^2.
$$
By the way, if $ab \mid (a+b)^2$, you can show that $a=b$ (see here).
