Find three sets $A$, $B$, $C$, each of them non-empty, such that $(A\cap B)\cup C=A\cap(B\cup C)$ and $(A\cap B)\cup C\neq A\cap(B\cup C)$ I need to find three sets for both statements I have above. I have tried drawing Venn diagrams and shading appropriately then adding numbers in each shaded region to try and guess and check for sets $A$, $B$ and $C$ but am unsure how to find an appropriate solution.


Consider the Venn diagram above. Let $$\begin{align*}A&=\text{Yellow}\cup \text{Green}\cup \text{Red}\cup \text{Dark},\\ B&=\text{Cyan}\cup \text{Blue}\cup \text{Green}\cup \text{Dark},\\ C&=\text{Magenta}\cup \text{Blue}\cup \text{Red}\cup \text{Dark}\end{align*}$$ then:
$$A\cap B=\text{Green}\cup \text{Dark}\\
(A\cap B)\cup C=\text{Magenta}\cup \text{Blue}\cup \text{Green}\cup \text{Red}\cup \text{Dark}\\
B\cup C=\text{Cyan}\cup \text{Magenta}\cup \text{Blue}\cup \text{Green}\cup \text{Red}\cup \text{Dark}\\
A\cap (B\cup C)=\text{Green}\cup \text{Red}\cup \text{Dark}$$
We see if $\text{Magenta}\cup \text{Blue}$ is empty, then equality holds, if not empty, the inequality holds.
In other words, it depends on whether $((A\cap B)\cup C)\setminus(A\cap (B\cup C))=C\setminus A$ is empty or not,
i.e. if $C\subseteq A$ then equality holds, if $C\not\subseteq A$ the inequality holds.
From here the examples (if not to consider the diagram above as the example of sets of points on a plane) may be derived rather fast, e.g. 
$$A=\{1,2\},\,B=\{3\},\,C=\{2\}\hbox{ for the first case and}$$
$$A=\{1\},\,B=\{3\},\,C=\{2\}\hbox{ for the second one.}$$
