$AB=BA$, $AC=CA$ we need to show $BC=CB$ $A,B,C$ are three $n\times n$ matrices such that $A$ has $n$ distinct eigenvalues, $AB=BA$, and $AC=CA$.   We need to show $BC=CB$.
Let $A$ be diagonalizable with eigenvectors say, $v_1,\dots,v_n$, so there exists $P$ such that $PAP^{-1}=D$. What more I can say from the information? Please help.

First note: For any matrices $X$ and $Y$ and invertible matrix $P$, the matrices $X$ and $Y$ commute if and only if the matrices $PXP^{-1}$ and $PYP^{-1}$ commute.
So if $PAP^{-1} = D$ then $D$ commutes with $B' = PBP^{-1}$ and $C' = PCP^{-1}$.  You know that $A$ has $n$ distinct eigenvalues therefore there are $n$ distinct entries on the diagonal of $D$.  Use this to show that any matrix that commutes with $D$ must itself be diagonal.  Then note that diagonal matrices commute therefore $B'$ and $C'$ commute, therefore $B$ and $C$ commute.
