Example of commutative ring that doesn't satisfy distribution of intersection over addition 
I'm trying to find an example of commutative ring $R$ and ideals $\mathfrak a,\mathfrak b,\mathfrak c \in R$ such that
  $$\mathfrak a \cap (\mathfrak b + \mathfrak c) \neq \mathfrak a \cap \mathfrak b +\mathfrak a \cap \mathfrak c .$$  

I know that $R \neq \mathbb Z, \mathbb Z_n$ and none of $\mathfrak a,\mathfrak b,\mathfrak c$ is contained into another else.
I've tried  $\mathbb Z[x]$ and other rings but i couldn't find any example. 
 A: In $\mathbb{C}[x,y]$ we have $(x)\cap(x+y,x-y) = (x)\cap(x,y) = (x)$, but $(x^2+xy)+(x^2-xy) = (x^2,xy)$. 
A: Hint $\  $ Distributivity easily yields that a finitely generated ideal $\,=1\,$ if it contains a cancellable element $\rm\:z\:$ that is $\rm\:lcm$-coprime to the generators, e.g. for a $2$-generated ideal $\rm\:(x,y)$
$$\rm\ \begin{array}{} (x)\cap(z)\ =\ (xz)\\ \rm (y)\cap(z)\ =\ (yz)\end{array}\ \ \ and\ \ \ (x,y)\supseteq (z)\ \ \Rightarrow\ \ (x,y) = 1$$
Proof $\rm\ \ (z) = (x,y)\cap (z) = (x)\cap(z) + (y)\cap(z) = (x,y)z\:$ so $\rm\:(x,y)=1\:$ via cancel $\rm\,z.$
Now examples are clear, e.g. let $\rm\ z = x\!+\!y\ $ for $\rm\ x,y \in \mathbb Q[x,y],\: $ or $\rm\ x\,$ and $\rm\,y=2\in \mathbb Z[x].$
A: In $\mathbb Z[x]$ you can try the following example: $\mathfrak a=(x)$, $\mathfrak b=(x-2)$ and $\mathfrak c=(x+2)$.

 We have $\mathfrak b+\mathfrak c=(4,x+2)$, $\mathfrak a\cap\mathfrak b+\mathfrak a\cap\mathfrak c=(x(x-2),x(x+2))$, $2x\in\mathfrak a\cap(\mathfrak b+\mathfrak c)$, and $2x\notin(x(x-2),x(x+2))$.

