To prove the mutual independence of events If A$,B$ and $C$ are random events in a sample space and if $A,
B$ and $C$ are pairwise independent and $A$ is independent of $(B \cup C)$, then is it true that $A,B$ and $C$ are mutually independent.
My Attempt : (with questions of this type at my college the answer is usually in the affirmative.)
So to prove that  $A,B$ and $C$ are mutually independent, all that remains to show is that $P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$. We know that $P(A \cap (B \cup C))=P(A) \times P(B \cup C)$. How do I go about this? 
 A: You have correctly identified the goal : We know about all necessary equations for mutual independence, except the final one : $P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$.
Why not directly start from $P(A \cap (B \cup C))=P(A) \times P(B \cup C)$ ? From there you can probably apply the union rule to the right-hand side :
$P(B \cup C)=P(B) + P(C) - P(B \cap C)$
And on the left-hand side you can transform $A \cap (B \cup C)$ into $(A\cap B)\cup(A\cap C)$ and apply the union rule again.
Where do you land from there ?
PS : For those reading but not knowing : Mutual independence between events require that all possible intersections between events follow the "multiplication rule", similar to the one we find in pairwise independence :
$P(A\cap B) = P(A) \times P(B) \\
P(A\cap C) = P(A) \times P(C) \\
P(B\cap C) = P(B) \times P(C) \\
P(A \cap B \cap C) = P(A) \times P(B) \times P(C)$
The first three are already known from pairwise independence in the problem statement.
A: $$P(A\cap (B\cup C))=P(A\cap B)+P(A\cap C)-P(A\cap B\cap C)\\\implies P(A)(P(B)+P(C)-P(B\cap C))=P(A)(P(B)+P(C))-P(A\cap B\cap C)\\\implies P(A)P(B)P(C)=P(A\cap B\cap C)$$
