Calculating $E[(X-E[X])^3]$ by mgf 
Calculate by mgf $E[(X-E[X])^3]$ where
a. $X\sim B(n,p)$
b.$X\sim N(\mu,\sigma)$

Before I begin I thought symbolizing $Y=X-E[X]$ and then I'd derivative $M_Y(t)$ three times substitute $t=0$ and solve both questions but I'm not sure about the distribution of Y.
My question is "Does subtracting a constant changes the distribution of random variable $\sim B(n,p)\text{ or }  N(\mu,\sigma)$"?
EDIT:
I got $np(2p−1)(p−1)$ in binomial and 0 in normal. The next question is why the binomial tend to 0 as $n→∞$?


My question is "Does subtracting a constant changes the distribution $B(n,p)$ or $N(\mu,\sigma)$"?

Well, of course it does... For example, $X$ is $N(\mu,\sigma)$ iff $X+c$ is $N(\mu+c,\sigma)$, hence $X-E[X]$ is $N(0,\sigma)$.
But this is not what you need to solve the exercise. Rather, you should first compute $E[X]$ and $M_X(t)=E[\mathrm e^{tX}]$, then $M_Y(t)=E[\mathrm e^{tY}]$. This last step is easy since $Y=X-E[X]$ hence $M_Y(t)=\mathrm e^{-E[X]t}M_X(t)$. Can you proceed from there?
