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 $n→∞$?

Yes, it changes the distribution. For one thing, the mean changes by that constant. In the binomial case, the distribution is no longer binomial. In the normal case, the new distribution is normal, mean $\mu-c$, variance $\sigma^2$, where $c$ is the constant you subtracted.  
We will look at the problem in two ways. The second way, which is the better way, uses the fact that the mgf of $X-c$ is a close relative of the mgf of $X$.
First way: One perhaps slightly painful but mechanical way to find the expectation of $(X-E(X))^3$ is to expand the cube. For simplicity write $\mu$ for $E(X)$. So we want $E(X^3-3\mu X^2+3\mu^2X-\mu^3)$. By the linearity of expectation, the mean of this expanded object is 
$$E(X^3)-3\mu E(X^2)+3\mu^2 E(X)-\mu^3.$$
Now all the missing bits can be picked up from the mgf of $X$.
Second way: Let $Y=X-\mu$, where $\mu=E(X)$.  Recall that the mgf of $Y$ is $E(e^{tY})$. This is $E(e^{t(X-\mu)})$, which is $e^{-t\mu} E(e^{tX})$.
We have found that the moment generating function of $Y=X-\mu$ is $e^{-\mu t}$ times the moment generating function of $X$.
Now for your two problems do this (i) Write down the mgf of $X$; (ii) Multiply by $e^{-\mu t}$. Now you have the mgf of $Y$. You can read off $E(Y^3)$ from the moment generating function of $Y$. 
For the normal, the answer you get should not come as a surprise, since $Y$ is normal with mean $0$ and variance $\sigma^2$. 
