$A\unlhd G$ , $B\unlhd G$ and $C\unlhd G$ then $A(B∩C)$ is a normal subgroup of $G$ If $A$ normal to $G$ , $B$ normal to $G$ and $C$ normal to $G$ then how can I show that$$A(B∩C)\unlhd G$$ 
how can i solve this problem? Thanks!

You know that if $B,C$ be subgroups of a group so does their intersection. Moreover if one of subgroups $A$ and $B\cap C$ are normal in $G$, so we have a theorem saying $A(B\cap C)\leq G$ also. Now show that the normality of $A(B\cap C)$ in $G$. In fact, show that: $$\forall x\in A(B\cap C), g\in G$$ we have $g^{-1}xg\in A(B\cap C)$ as well where $g\in G$ is an arbitrary element.
