$\displaystyle given \ \ \ \ P(A\cap B\cap C)= p_{A}p_{B}p_{C} ,\ \ \ prove :\ \ \ P(A\cap B)=p_{A}p_{B} $
i try to set: $\displaystyle C=A\cap B, \ \ p_{C}=1 $
but unsure if thats right.
Try $\displaystyle AB=ABC\cup ABC'$
I assume you have independence between A, B and C, which implies independence between A, B and C'.
So $\displaystyle P(ABC)=P(A)P(B)P(C)$ and $\displaystyle P(ABC')=P(A)P(B)P(C')$
add these.... $\displaystyle P(A)P(B)P(C)+P(A)P(B)P(C')=P(A)P(B)\bigl[P(C)+P(C')\bigr]=P(AB)$