Using Binomial Distribution to evaluate this probability distribution I am trying to find the value of a skewed distribution but can't make sense of what to plug in to evaluate the answer. 
This is the given:
$$
\text{Let X be Binomial(n, p).  }
\text{Using that, evaluate:}
$$
$$
\beta = \frac{E[(X-\mu)^3]}{\sigma^3}
$$
Now, I expanded the numerator and got this: (wikipedia)
$$
\beta = \frac{E[X^3] - 3\mu\sigma^2 - \mu^3}{\sigma^3}
$$
and I know that $\mu=np$ and $\sigma=\sqrt{3np(1-p)}$ and this is what it simplifies to from what I did
$$
\beta = \frac{E[X^3] - 3np(3np(1-p)) - (np)^3}{(\sqrt{3np(1-p)})^3}
$$
The Problem
The issue is that I can't make sense of $E[X^3]$. I don't know how to evaluate that in order to get a numerical value of $\beta$ for arbitrary n and p values. The binomial distribution should take arguments x, n and p right? What is x here?
Would $E[X^3]$ be just $(np)^3$?
Thanks
 A: The Pochhammer symbol $(x)_k$ is used by some people, perhaps especially combinatorialists, to represent the falling factorial:
$$
(x)_k = x(x-1)(x-2)\cdots(x-k+1)
$$
and by others, perhaps especially those who work with special functions, to represent the rising factorial
$$
(x)_k = x(x+1)(x+2)\cdots(x+k-1).
$$
I will follow the former convention.  One way to find $E(X^3)$ is first to write $X^3$ as a linear combination of falling factorials:
$$
X^3 = (X)_3 + 3(X)_2 + X.
$$
Then
$$
E((X)_3) = \sum_{x=0}^n (x)_3 \binom n x p^x (1-p)^{n-x} = \sum_{x=0}^n x(x-1)(x-2)\frac{n!}{(n-x)!x!} p^x (1-p)^{n-x}
$$
$$
= \sum_{x=3}^n x(x-1)(x-2)\frac{n!}{(n-x)!x!}  p^x (1-p)^{n-x}
$$
Notice we're now starting at $x=3$ rather than $x=0$, since the terms for $0,1,2$ are $0$.  Now do a substitution: let $y=x-3$ and notice that as $x$ goes from $3$ to $n$ then $y$ goes from $0$ to $n-3$:
$$
\sum_{y=0}^{n-3} (y+3)(y+3)(y+1) \binom{n!}{((n-3)-y)!(y+3)!} p^{y+3} (1-p)^{(n-3)-y}
$$
$$
= \sum_{y=0}^{n-3} n(n-1)(n-3) \frac{(n-3)!}{((n-3)-y)!y!} p^{y+3} (1-p)^{(n-3)-y}
$$
$$
= (n)_3 p^3 \sum_{y=0}^{n-3} \binom{n-3}{y} p^y (1-p)^{(n-3)-y}
$$
$$
= (n)_3 p^3 \cdot 1.
$$
Then do similar things with $E((X)_2)$, and so on.
A: If we are interested in higher moments, the moment generating function $M_X(t)$ of $X$ can be useful. This is given by 
$$M_X(t)=E(e^{tX}=\sum_{x=0}^n e{tx}\binom{n}{x}p^xq^{n-x},$$
 where $q=1-p$. The moment generating function can be rewritten as 
$$\sum_{x=0}^n \binom{n}{x}(pe^t)^xq^{n-x},$$
which we recognize as $(pe^t+q)^n$.
Expanding $E(e^{tX})$ as a power series, we get
$$M_X(t)=1+tE(X)+\frac{t^2}{2!}E(X^2)+\frac{t^3}{3!}E(X^3)+\cdots.$$
Differentiate $M_X(t)$ three times, and set $t=0$. We get 
$$E(X^3)=M_X'''(0).$$ 
The differentiation three times of $(pe^t+q)^n$ is straightforward.
A: Recall the definition of the expectation:
$$
     \mathbb{E}\left(X^3\right) = \sum_{k=0}^n k^3 \mathbb{P}\left(X=k\right) = \sum_{k=0}^n k^3 \binom{n}{k} p^k (1-p)^{n-k} \tag{1}
$$
In order to evaluate $\mathbb{E}\left(X^3\right)$ it is useful to consider a different sum instead. Consider
$$
    \mathcal{P}_X(z) = \sum_{k=0}^n z^k \mathbb{P}\left(X=k\right)
$$
known as the probability generating function. Then
$$
    \mathbb{E}\left(X\right) = \left.z \frac{\mathrm{d}}{\mathrm{d}z} \mathcal{P}_X(z)\right|_{z=1}, \quad \mathbb{E}\left(X^2\right) = \left. z \frac{\mathrm{d}}{\mathrm{d}z} z \frac{\mathrm{d}}{\mathrm{d}z} \mathcal{P}_X(z)\right|_{z=1}, \quad \mathbb{E}\left(X^3\right) = \left. z \frac{\mathrm{d}}{\mathrm{d}z} z \frac{\mathrm{d}}{\mathrm{d}z} z \frac{\mathrm{d}}{\mathrm{d}z} \mathcal{P}_X(z)\right|_{z=1}
$$
Evaluation of $\mathcal{P}_X(z)$ for the binomial distribution is especially easy:
$$\begin{eqnarray}
  \mathcal{P}_X(z) &=& \sum_{k=0}^n z^k \mathbb{P}\left(X=k\right) = \sum_{k=0}^n z^k \binom{n}{k} p^k (1-p)^{n-k} = \sum_{k=0}^n \binom{n}{k} (z p)^k (1-p)^{n-k} \\
  &=& \left(1-p + p z\right)^n
\end{eqnarray}
$$
Instead of computing $\mathbb{E}(X^3)$, it is easier to evaluate
$$
  \left. \frac{\mathrm{d}^3}{\mathrm{d}z^3} \mathcal{P}_X(z)\right|_{z=1} = \mathbb{E}\left(X(X-1)(X-2)\right) = n(n-1)(n-2) p^3
$$
Now, after some algebra we get:
$$
  \mathbb{E}\left(X^3\right) = n p \left( 1 + 3 (n-1)p  + (n-1)(n-2) p^2\right)
$$
